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Solution - of - Linear - System
- Theory - and - Design - 3ed -
for - Chi - Tsong - Chen
Design
Centro Federal de Educação Tecnológica de Minas Gerais (CEFET/MG)
106 pag.
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Linear System Theory and Design SA01010048 LING QING
1
2.1 Consider the memoryless system with characteristics shown in Fig 2.19, in which u denotes
the input and y the output. Which of them is a linear system? Is it possible to introduce a new
output so that the system in Fig 2.19(b) is linear?
Figure 2.19
Translation: 考虑具 图 2.19 中表 性 记忆 统。其中 u 表 输入，y 表 输出。
下 哪一个 性 统？可以找到一个 输出，使得图 2.19(b)中 统 性
吗？
Answer: The input-output relation in Fig 2.1(a) can be described as:
uay *=
Here a is a constant. It is a memoryless system. Easy to testify that it is a linear system.
The input-output relation in Fig 2.1(b) can be described as:
buay += *
Here a and b are all constants. Testify whether it has the property of additivity. Let:
buay += 11 *
buay += 22 *
then:
buuayy *2)(*)( 2121 ++=+
So it does not has the property of additivity, therefore, is not a linear system.
But we can introduce a new output so that it is linear. Let:
byz −=
uaz *=
z is the new output introduced. Easy to testify that it is a linear system.
The input-output relation in Fig 2.1(c) can be described as:
uuay *)(=
a(u) is a function of input u. Choose two different input, get the outputs:
111 *uay =
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Linear System Theory and Design SA01010048 LING QING
2
222 *uay =
Assure:
21 aa ≠
then:
221121 **)( uauayy +=+
So it does not has the property of additivity, therefore, is not a linear system.
2.2 The impulse response of an ideal lowpass filter is given by
)(2
)(2sin
2)(
0
0
tt
tt
tg
−
−
=
ω
ω
ω
for all t, where w and to are constants. Is the ideal lowpass filter causal? Is is possible to built
the filter in the real world?
Translation: 想低通 器 冲 响应如式所 。对于所 t，w 和 to，都 常 。
想低通 器 因 吗？ 实世 中 可能 造这 器吗？
Answer: Consider two different time: ts and tr, ts < tr, the value of g(ts-tr) denotes the output at
time ts, excited by the impulse input at time tr. It indicates that the system output at time
ts is dependent on future input at time tr. In other words, the system is not causal. We
know that all physical system should be causal, so it is impossible to built the filter in
the real world.
2.3 Consider a system whose input u and output y are related by
>
≤
==
atfor
atfortu
tuPty a
0
)(
:))(()(
where a is a fixed constant. The system is called a truncation operator, which chops off the
input after time a. Is the system linear? Is it time-invariant? Is it causal?
Translation: 考虑具 如式所 输入输出关 统，a 一个 定 常 。这个 统 作
截 器。它截 a 之后 输入。这个 统 性 吗？它 定常 吗？ 因
吗？
Answer: Consider the input-output relation at any time t, t<=a:
uy =
Easy to testify that it is linear.
Consider the input-output relation at any time t, t>a:
0=y
Easy to testify that it is linear. So for any time, the system is linear.
Consider whether it is time-invariable. Define the initial time of input to, system input is
u(t), t>=to. Let to<a, so It decides the system output y(t), t>=to:
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Linear System Theory and Design SA01010048 LING QING
3
≤≤
=
totherfor
attfortu
ty
0
)(
)(
0
Shift the initial time to to+T. Let to+T>a , then input is u(t-T), t>=to+T. System output:
0)(' =ty
Suppose that u(t) is not equal to 0, y’(t) is not equal to y(t-T). According to the definition,
this system is not time-invariant.
For any time t, system output y(t) is decided by current input u(t) exclusively. So it is a
causal system.
2.4 The input and output of an initially relaxed system can be denoted by y=Hu, where H is some
mathematical operator. Show that if the system is causal, then
uHPPHuPyP aaaa ==
where Pa is the truncation operator defined in Problem 2.3. Is it true PaHu=HPau?
Translation: 一个初始 弛 统 输入输出可以 述为：y=Hu，这里 H 学运 ，
说 假如 统 因 性 ， 如式所 关 。这里 Pa 2.3 中定义 截 函
。PaHu=HPau 吗？
Answer: Notice y=Hu, so:
HuPyP aa =
Define the initial time 0, since the system is causal, output y begins in time 0.
If a<=0,then u=Hu. Add operation PaH in both left and right of the equation:
uHPPHuP aaa =
If a>0, we can divide u to 2 parts:
≤≤
=
totherfor
atfortu
tp
0
0)(
)(
>
=
totherfor
atfortu
tq
0
)(
)(
u(t)=p(t)+q(t). Pay attention that the system is casual, so the output excited by q(t) can’t
affect that of p(t). It is to say, system output from 0 to a is decided only by p(t). Since
PaHu chops off Hu after time a, easy to conclude PaHu=PaHp(t). Notice that p(t)=Pau,
also we have:
uHPPHuP aaa =
It means under any condition, the following equation is correct:
uHPPHuPyP aaaa ==
PaHu=HPau is false. Consider a delay operator H, Hu(t)=u(t-2), and a=1, u(t) is a step
input begins at time 0, then PaHu covers from 1 to 2, but HPau covers from 1 to 3.
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Linear System Theory and Design SA01010048 LING QING
4
2.5 Consider a system with input u and output y. Three experiments are performed on the system
using the inputs u1(t), u2(t) and u3(t) for t>=0. In each case, the initial state x(0) at time t=0 is
the same. The corresponding outputs are denoted by y1,y2 and y3. Which of the following
statements are correct if x(0)<>0?
1. If u3=u1+u2, then y3=y1+y2.
2. If u3=0.5(u1+u2), then y3=0.5(y1+y2).
3. If u3=u1-u2, then y3=y1-y2.
Translation: 考虑具 输入 u 输出 y 统。在 统上进行三 实验，输入分别为 u1(t),
u2(t) 和 u3(t)，t>=0。 情况下， 刻 初态 x(0)都 同 。 应 输出表
为 y1,y2 和 y3。在 x(0)不 于 情况下，下 哪 说 ？
Answer: A linear system has the superposition property:
02211
02211
022011
),()(
),()(
)()(
tttyty
tttutu
txtx
≥+→
≥+
+
αα
αα
αα
In case 1:
11 =α 12 =α
)0()0(2)()( 022011 xxtxtx ≠=+αα
So y3<>y1+y2.
In case 2:
5.01 =α 5.02 =α
)0()()( 022011 xtxtx =+αα
So y3=0.5(y1+y2).
In case 3:
11 =α 12 −=α
)0(0)()( 022011 xtxtx ≠=+αα
So y3<>y1-y2.
2.6 Consider a system whose input and outputare related by
=−
≠−−
=
0)1(0
0)1()1(/)(
)(
2
tuif
tuiftutu
ty
for all t.
Show that the system satisfies the homogeneity property but not the additivity property.
Translation: 考虑输入输出关 如式 统,证 统 足齐 性,但 不 足可加性.
Answer: Suppose the system is initially relaxed, system input:
)()( tutp α=
a is any real constant. Then system output q(t):
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Linear System Theory and Design SA01010048 LING QING
5
=−
≠−−
=
0)1(0
0)1()1(/)(
)(
2
tpif
tpiftptp
tq
=−
≠−−
=
0)1(0
0)1()1(/)(2
tuif
tuiftutuα
So it satisfies the homogeneity property.
If the system satisfies the additivity property, consider system input m(t) and n(t),
m(0)=1, m(1)=2; n(0)=-1, n(1)=3. Then system outputs at time 1 are:
4)0(/)1()1( 2 == mmr
9)0(/)1()1( 2 −== nns
0)]0()0(/[)]1()1([)1( 2 =++= nmnmy
）1()1( sr +≠
So the system does not satisfy the additivity property.
2.7 Show that if the additivity property holds, then the homogeneity property holds for all rational
numbers a . Thus if a system has “continuity” property, then additivity implies homogeneity.
Translation: 说 统如 具 可加性，那么对所 a 具 齐 性。因而对具
连续性质 统，可加性导致齐 性。
Answer: Any rational number a can be denoted by:
nma /=
Here m and n are both integer. Firstly, prove that if system input-output can be described
as following:
yx →
then:
mymx →
Easy to conclude it from additivity.
Secondly, prove that if a system input-output can be described as following:
yx →
then:
nynx // →
Suppose:
unx →/
Using additivity:
nuxnxn →=)/(*
So:
nuy =
nyu /=
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Linear System Theory and Design SA01010048 LING QING
6
It is to say that:
nynx // →
Then:
nmynmx /*/* →
ayax →
It is the property of homogeneity.
2.8 Let g(t,T)=g(t+a,T+a) for all t,T and a. Show that g(t,T) depends only on t-T.
Translation: 设对于所 t,T 和 a，g(t,T)=g(t+a,T+a)。说 g(t,T)仅依赖于 t-T。
Answer: Define:
Ttx += Tty −=
So:
2
yx
t
+
=
2
yx
T
−
=
Then:
)
2
,
2
(),(
yxyx
gTtg
−+
=
)
2
,
2
( a
yx
a
yx
g +
−
+
+
=
)
22
,
22
(
yxyxyxyx
g
+−
+
−+−
+
+
=
)0,(yg=
So:
0
)0,(),(
=
∂
∂
=
∂
∂
x
yg
x
Ttg
It proves that g(t,T) depends only on t-T.
2.9 Consider a system with impulse response as shown in Fig2.20(a). What is the zero-state
response excited by the input u(t) shown in Fig2.20(b)?
Fig2.20
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Linear System Theory and Design SA01010048 LING QING
7
Translation: 考虑冲 响应如图 2.20(a)所 统， 如图 2.20(b)所 输入 u(t) 励
态响应 什么？
Answer: Write out the function of g(t) and u(t):
≤≤−
≤≤
=
212
10
)(
tt
tt
tg
≤≤−
≤≤
=
211
101
)(
t
t
tu
then y(t) equals to the convolution integral:
∫ −=
t
drrturgty
0
)()()(
If 0=<t=<1, 0=<r=<1, 0<=t-r<=1:
∫=
t
rdrty
0
)(
2
2t
=
If 1<=t<=2:
)(ty ∫
−
−=
1
0
)()(
t
drrturg ∫
−
−+
1
1
)()(
t
drrturg ∫ −+
t
drrturg
1
)()(
)(1 ty= )(2 ty+ )(3 ty+
Calculate integral separately:
)(1 ty ∫
−
−=
1
0
)()(
t
drrturg 10 ≤≤ r 21 ≤−≤ rt
∫
−
−=
1
0
t
rdr
2
)1( 2−−
=
t
)(2 ty ∫
−
−=
1
1
)()(
t
drrturg 10 ≤≤ r 10 ≤−≤ rt
∫
−
=
1
1t
rdr
2
)1(
2
1 2−
−=
t
)(3 ty ∫ −=
t
drrturg
1
)()( 21 ≤≤ r 10 ≤−≤ rt
∫ −=
t
drr
1
)2(
2
1
)1(2
2 −
−−=
t
t
)(ty )(1 ty= )(2 ty+ )(3 ty+ 24
2
3 2 −+−= tt
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Linear System Theory and Design SA01010048 LING QING
8
2.10 Consider a system described by
uuyyy −=−+
••••
32
What are the transfer function and the impulse response of the system?
Translation: 考虑如式所 述 统，它 传递函 和冲 响应 什么？
Answer: Applying the Laplace transform to system input-output equation, supposing that the
System is initial relaxed:
)()()(3)(2)(2 sYssYsYssYsYs −=−+
System transfer function:
3
1
32
1
)(
)(
)(
2 +
=
−+
−
==
sss
s
sY
sU
sG
Impulse response:
te
s
LsGLtg 311 ]
3
1
[)]([)( −−− =
+
==
2.11 Let y(t) be the unit-step response of a linear time-invariant system. Show that the impulse
response of the system equals dy(t)/dt.
Translation: y(t) 性定常 统 单位 跃响应。说 统 冲 响应 于 dy(t)/dt.
Answer: Let m(t) be the impulse response, and system transfer function is G(s):
s
sGsY
1
*)()( =
)()( sGsM =
ssYsM *)()( =
So:
dttdytm /)()( =
2.12 Consider a two-input and two-output system described by
)()()()()()()()( 212111212111 tupNtupNtypDtypD +=+
)()()()()()()()( 222121222121 tupNtupNtypDtypD +=+
where Nij and Dij are polynomials of p:=d/dt. What is the transfer matrix of the system?
Translation: 考虑如式 述 两输入两输出 统， Nij 和 Dij p:=d/dt 多 式。 统
传递 什么？
Answer: For any polynomial of p, N(p), its Laplace transform is N(s).
Applying Laplace transform to the input-output equation:
)()()()()()()()( 212111212111 sUsNsUsNsYsDsYsD +=+
)()()()()()()()( 222121222121 sUsNsUsNsYsDsYsD +=+
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Linear System Theory and Design SA01010048 LING QING
9
Write to the form of matrix:
)(
)(
)()(
)()(
2
1
2221
1211
sY
sY
sDsD
sDsD
=
)(
)(
)()(
)()(
2
1
2221
1211
sU
sU
sNsN
sNsN
)(
)(
2
1
sY
sY
=
1
2221
1211
)()(
)()(
−
sDsD
sDsD
)(
)(
)()(
)()(
2
1
2221
1211
sU
sU
sNsN
sNsN
So the transfer function matrix is:
)(sG =
1
22211211
)()(
)()(
−
sDsD
sDsD
)()(
)()(
2221
1211
sNsN
sNsN
By the premising that the matrix inverse:
1
2221
1211
)()(
)()(
−
sDsD
sDsD
exists.
2.11 Consider the feedback systems shows in Fig2.5. Show that the unit-step responses of the
positive-feedback system are as shown in Fig2.21(a) for a=1 and in Fig2.21(b) for a=0.5.
Show also that the unit-step responses of the negative-feedback system are as shown in Fig
2.21(c) and 2.21(d), respectively, for a=1 and a=0.5.
Fig 2.21
Translation: 考虑图 2.5 中所 反馈 统。说 反馈 统 单位 跃响应，当 a=1 ，如
图 2.21(a)所 。当 a=0.5 ，如图 2.21(b)所 。说 负反馈 统 单位 跃响应
如图 2.21(c)和 2.21(b)所 ， 应地，对 a=1 和 a=0.5。
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Linear System Theory and Design SA01010048 LING QING
10
Answer: Firstly, consider the positive-feedback system. It’s impulse response is:
∑
∞
=
−=
1
)()(
i
i itatg d
Using convolution integral:
∑
∞
=
−=
1
)()(
i
i itraty
When input is unit-step signal:
∑
=
=
n
i
iany
1
)(
)()( nyty = 1+≤≤ ntn
Easy to draw the response curve, for a=1 and a=0.5, respectively, as Fig 2.21(a) and Fig
2.21(b) shown.
Secondly, consider the negative-feedback system. It’s impulse response is:
∑
∞
=
−−−=
1
)()()(
i
i itatg d
Using convolution integral:
∑
∞
=
−−−=
1
)()()(
i
i itraty
When input is unit-step signal:
∑
=
−−=
n
i
iany
1
)()(
)()( nyty = 1+≤≤ ntn
Easy to draw the response curve, for a=1 and a=0.5, respectively, as Fig 2.21(c) and Fig
2.21(d) shown.
2.14 Draw an op-amp circuit diagram for
uxx
−
+
−
=
•
4
2
50
42
[ ] uxy 2103 −=
2.15 Find state equations to describe the pendulum system in Fig 2.22. The systems are useful to
model one- or two-link robotic manipulators. If θ , 1θ and 2θ are very small, can you
consider the two systems as linear?
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Linear System Theory and Design SA01010048 LING QING
11
Translation: 试找出图 2.22 所 单 统 态 。这个 统对 一个或两个连接
器人 作臂很 。假如角度都很小 ，能否考虑 统为 性？
Answer: For Fig2.22(a), the application of Newton’s law to the linear movements yields:
)cossin()cos(cos
2
2
2
θθθθθθ
•••
−−==− mll
dt
d
mmgf
)sincos()sin(sin
2
2
2
θθθθθθ
•••
−==− mll
dt
d
mfu
Assuming θ and
•
θ to be small, we can use the approximation θsin =θ , θcos =1.
By retaining only the linear terms in θ and
•
θ , we obtain mgf = and:
u
mll
g 1
+−=
••
θθ
Select state variables as θ=1x ,
•
=θ2x and output θ=y
u
ml
x
lg
x
+
−
=
•
/1
1
0/
10
[ ]xy 01=
For Fig2.22(b), the application of Newton’s law to the linear movements yields:
)cos(coscos 112
2
112211 θθθ l
dt
d
mgmff =−−
)cossin( 1
2
11111 θθθθ
•••
−−= lm
)sin(sinsin 112
2
11122 θθθ l
dt
d
mff =−
)sincos( 1
2
11111 θθθθ
•••
−= lm
)coscos(cos 22112
2
2222 θθθ ll
dt
d
mgmf +=−
)cossin( 1
2
11112 θθθθ
•••
−−= lm )cossin( 2
2
22222 θθθθ
•••
−−+ lm
)sinsin(sin 22112
2
222 θθθ ll
dt
d
mfu +=−
)sincos( 1
2
11112 θθθθ
•••
−= lm )sincos( 2
2
22222 θθθθ
•••
−+ lm
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Linear System Theory and Design SA01010048 LING QING
12
Assuming 1θ , 2θ and 1
•
θ , 2
•
θ to be small, we can use the approximation 1sinθ = 1θ ,
2sinθ = 2θ , 1cosθ =1, 2cosθ =1. By retaining only the linear terms in 1θ , 2θ and
1
•
θ , 2
•
θ , we obtain gmf 22 = , gmmf )( 211 += and:
2
11
2
1
11
21
1
)(
θθθ
lm
gm
lm
gmm
+
+
−=
••
u
lmlm
gmm
lm
gmm
22
2
21
21
1
21
21
2
1)()(
+
+
−
+
=
••
θθθ
Select state variables as 11 θ=x , 12
•
=θx , 23 θ=x , 24
•
=θx and output
=
2
1
2
1
θ
θ
y
y
:
+−+
+−
=
•
•
•
•
4
3
2
1
22212121
1121121
4
3
2
1
0/)(0/)(
1000
0/0/)(
0010
x
x
x
x
lmgmmlmgmm
lmgmlmgmm
x
x
x
x
+ u
lm
22/1
0
0
0
=
0100
0001
2
1
y
y
4
3
2
1
x
x
x
x
2.17 The soft landing phase of a lunar module descending on the moon can be modeled as shown
in Fig2.24. The thrust generated is assumed to be proportional to the derivation of m, where
m is the mass of the module. Then the system can be described by
mgmkym −−=
•••
Where g is the gravity constant on the lunar surface. Define state variables of the system as:
yx =1 ,
•
= yx2 , mx =3 ,
•
= my
Find a state-space equation to describe the system.
Translation: 舱 落在 ，软 型如图 2.24 所 。产 冲 力与 m
微分成 。 统可以 述如式所 形式。g 表 重力加速度常 。定
义 态变 如式所 ，试图找出 统 态 述。
Answer: The system is not linear, so we can linearize it.
Suppose:
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Linear System Theory and Design SA01010048 LING QING
13
−
+−= ygty 2/2
−
••
+−= ygty
−
••••
+−= ygy
−
+= mmm 0
−
••
= mm
So:
)( 0
−
+ mm
−
••
+− )( yg =
−
•
− mk gmm )( 0
−
+−
+−−
−
gmgm0
−
••
=ym0
−
•
− mk gmgm
−
−− 0
−
••
=ym0
−
•
− mk
Define state variables as:
−−
= yx1 ,
−
•−
= yx 2 ,
−−
= mx3 ,
−
•−
= my
Then:
•
−
•
−
•
−
3
2
1
x
x
x
=
000
000
010
−
−
−
3
2
1
x
x
x
+
−
1
/
0
0mk
−
u
−
y = [ ]001
−
−
−
3
2
1
x
x
x
2.19 Find a state equation to describe the network shown in Fig2.26.Find also its transfer function.
Translation: 试写出 述图 2.26 所 网络 态 ，以及它 传递函 。
Answer: Select state variables as:
1x : Voltage of left capacitor
2x : Voltage of right capacitor
3x : Current of inductor
Applying Kirchhoff’s current law:
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Linear System Theory and Design SA01010048 LING QING
143x = RxLx /)( 32
•
−
RxxCu 11+=
•
32 xxCu +=
•
2xy =
From the upper equations, we get:
CuCRxx //11 +−=
•
ux +−= 1
CuCxx //32 +−=
•
ux +−= 3
LRxLxx // 323 −=
•
32 xx −=
2xy =
They can be combined in matrix form as:
•
•
•
3
2
1
x
x
x
=
−
−
−
110
100
001
3
2
1
x
x
x
+ u
0
1
1
[ ]
=
3
2
1
010
x
x
x
y
Use MATLAB to compute transfer function. We type:
A=[-1,0,0;0,0,-1;0,1,-1];
B=[1;1;0];
C=[0,1,0];
D=[0];
[N1,D1]=ss2tf(A,B,C,D,1)
Which yields:
N1 =
0 1.0000 2.0000 1.0000
D1 =
1.0000 2.0000 2.0000 1.0000
So the transfer function is:
122
12
)(
23
2^
+++
++
=
sss
ss
sG
1
1
2 ++
+
=
ss
s
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Linear System Theory and Design SA01010048 LING QING
15
2.20 Find a state equation to describe the network shown in Fig2.2. Compute also its transfer
matrix.
Translation: 试写出 述图 2.2 所 网络 态 ，计 它 传递函 。
Answer: Select state variables as Fig 2.2 shown. Applying Kirchhoff’s current law:
1u = 323121111 xRxLxxxCR ++++
••
22211
••
=+ xCuxC
223
•
= xCx
3231212311 )( xRxLxxuxRu ++++−=
•
3231 xRxLy +=
•
From the upper equations, we get:
12131 // CuCxx −=
•
232 / Cxx =
•
12111131112113 ///)(// LuRLuLxRRLxLxx +++−−−=
•
2113121 uRuxRxxy ++−−−=
They can be combined in matrix form as:
•
•
•
3
2
1
x
x
x
=
+−−− 12111
2
1
/)(
/100
/100
LRRLL
C
C
3
2
1
x
x
x
+
−
2
1
11
1
1 /
0
/1
/1
0
0
u
u
LR
C
L
[ ]
−−−=
3
2
1
111
x
x
x
Ry + [ ]
2
1
21
u
u
R
Applying Laplace Transform to upper equations:
)(
/1/1
)( 1
^
21111
21
^
su
sCsCRRsL
RsL
sy
++++
+
=
)(
/1/1
)/1)((
2
^
21111
2121 su
sCsCRRsL
sCRRsL
++++
++
+
++++
+
=
sCsCRRsL
RsL
sG
21111
21
^
/1/1
)(
++++
++
sCsCRRsL
sCRRsL
21111
2121
/1/1
)/1)((
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Linear System Theory and Design SA01010048 LING QING
16
2.18 Find the transfer functions from u to 1y and from 1y to y of the hydraulic tank system
shown in Fig2.25. Does the transfer function from u to y equal the product of the two
transfer functions? Is this also true for the system shown in Fig2.14?
Translation: 试写出图 2.25 所 统从u 到 1y 传递函 和从 1y 到 y 传递函 。从
u 到 y 传递函 于两个传递函 乘 吗？这对图 2.14 所 统也 吗？
Answer: Write out the equation about u , 1y and y :
111 / Rxy =
222 / Rxy =
dtyudxA /)( 111 −=
dtyydxA /)( 122 −=
Applying Laplace transform:
)1/(1/ 11
^
1
^
sRAuy +=
)1/(1/ 221
^^
sRAyy +=
)1)(1/(1/ 2211
^^
sRAsRAuy ++=
So:
=
^^
/ uy )/(
^
1
^
uy )/(
1
^^
yy
But it is not true for Fig2.14, because of the loading problem in the two tanks.
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3.1consider Fig3.1 ,what is the representation of the vector x with respect to the basis ( )21 iq ?
What is the representation of 1q with respect ro ( )22 qi ?图 3.1 中,向 x 关于 ( )21 iq 表
什么? 1q 关于 ( )22 qi 表 什么?
If we draw from x two lines in parallel with 2i and 1q , they tutersect at 1
3
1
q and 2
3
8
i as
shown , thus the representation of x with respect to the basis ( )21 iq is
′
3
8
3
1
, this can
be verified from [ ]
=
=
=
3
8
3
1
11
03
3
8
3
1
3
1
22 iqx
To find the representation of 1q with respect to ( )22 qi ,we draw from 1q two lines in
parallel with 2q and 2i , they intersect at -2 2i and 2
2
3
q , thus the representation of 1q
with respect to ( )22 qi , is
′
−
2
3
2 , this can be verified from
[ ]
−
=
−
=
=
2
3
2
21
20
2
3
2
1
3
221 qiq
3.2 what are the 1-morm ,2-norm , and infinite-norm of the vectors [ ] [ ]′=′−= 111,132 21 xx , 向
[ ] [ ]′=′−= 111,132 21 xx 1-范 ,2-范 和 −∞ 范 什么?
1max3max
3)111(14)1)3(2(
31116132
2211
2
1
222
2
2
2
1
222
1
'
1
2
1
1
2
3
1
1
1
1
====
=++==+−+==
=++==++==
∞∞
=
∑
iiii
i
i
xxxxxx
xxxx
xxx
3.3 find two orthonormal vectors that span the same space as the two vectors in problem 3.2 , 与 3.2 中 两个向
张成同一 两个 准 交向 .
Schmidt orthonormalization praedure ,
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[ ] [ ]
[ ]′===−=
′−==′−==
111
3
1
)(
132
14
1
132
2
2
2212
'
122
1
1
111
u
u
qxqxqxu
u
u
qxu
The two orthomormal vectors are
=
−=
1
1
1
3
1
1
3
2
14
1
21 qq
In fact , the vectors 1x and 2x are orthogonal because 01221 =′=′ xxxx so we can only
normalize the two vectors
==
−==
1
1
1
3
1
1
3
2
14
1
2
2
2
1
1
1
x
x
q
x
x
q
3.4 comsider an mn× matrix A with mn ≥ , if all colums of A are orthonormal , then
mIAA =′ , what can you say abort AA ′ ? 一个 mn× A( mn ≥ ), 如 A 所 列都
准 交 ,则 mIAA =′ AA ′ 怎么 ?
Let [ ] [ ]
mnijm
aaaaA
×
== 21 , if all colums of A are orthomormal , that is
∑
=
=
≠
==
n
i
liliii
jiif
jiif
aaaa
1
'
1
0
then
[ ]
[ ]
=≠
≠≠
=
==
=′
=
=
=
=′
∑
∑∑
=
×
==
jiif
jiif
aageneralin
aaaa
a
a
a
aaaAA
I
aaaaaa
aaaaaa
aaa
a
a
a
AA
jl
m
i
il
nnjl
m
i
il
m
i
ii
m
mi
m
mmmm
m
mi
m
1
0
)(
100
010
001
1
11
'
'
'
2
'
1
''
2
'
'
2
'
1
'
'
12
'
11
'
1
''
2
'
'
'
2
'
1
if A is a symmetric square matrix , that is to say , n=m jlil aa for every nli 2,1, =
then mIAA =′
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3.5 find the ranks and mullities of the following matrices 下列 和化 度
−−=
−
=
=
1000
2210
4321
011
023
114
100
000
010
321 AAA
134)(033)(123)(
3)(3)(2)(
321
321
=−==−==−=
===
ANullityANullityANullity
ArankArankArank
3.6 Find bases of the range spaces of the matrices in problem 3.5 3.5 中 值域 基
the last two columns of 1A are linearly independent , so the set
10
0
,
0
0
1
can be used as a
basis of the range space of 1A , all columns of 2A are linearly independent ,so the set
−
0
0
1
1
2
2
1
3
4
can be used as a basis of the range spare of 2A
let [ ]43213 aaaaA = ,where ia denotes of 3A 4321 aandaandaanda are
linearly independent , the third colums can be expressed as 213 2aaa +−= , so the set
{ }321 aaa can be used as a basis of the range space of 3A
3.7 consider the linear algebraic equation xx −
=
−
−
−
1
0
1
21
33
12
it has three equations and two
unknowns does a solution x exist in the equation ? is the solution unique ? does a solution exist if
[ ]′= 111y ?
性代 xx −
=
−
−
−
1
0
1
21
33
12
三个 两个 , 否 解?若 解 否
唯一?若 [ ]′= 111y 否 解?
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Let [ ],
21
33
12
21 aaA =
−
−
−
= clearly 1a and 2a are linearly independent , so rank(A)=2 ,
y is the sum of 1a and 2a ,that is rank([A y ])=2, rank(A)= rank([A y ]) so a solution x
exists in A x = y
Nullity(A)=2- rank(A)=0
The solution is unique [ ]′= 11x if [ ]′= 111y , then rank([A y ])=3≠ rank(A), that is
to say there doesn’t exist a solution in A x = y
3.8 find the general solution of
=
−−
1
2
3
1000
2210
4321
x how many parameters do you have ?
=
−−
1
2
3
1000
2210
4321
x 同解,同解中 了几个参 ?
Let
=
−−=
1
2
3
1000
2210
4321
yA we can readily obtain rank(A)= rank([A y ])=3 so
this y lies in the range space of A and [ ]′−= 101 ox p is a solution Nullity(A)=4-3=1 that
means the dimension of the null space of A is 1 , the number of parameters in the general solution
will be 1 , A basis of the null space of A is [ ]′−= on 121 thus the general solution of
A x = y can be expressed an
−
+
−
=+=
0
1
2
1
1
0
0
1
ααnxx p for any real α
α is the only parameter
3.9 find the solution in example 3.3 that has the smallest Euclidean norm 例 3 中具 小
范 解,
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the general solution in example 3.3 is
−
−
−+
=
−
+
−
+
−
=
2
1
21
1
21
42
1
0
2
0
0
1
1
1
0
0
4
0
α
α
αα
α
ααx for
any real 1α and 2α
the Euclidean norm of x is
16168453
)()()42(
2121
2
2
2
1
2
2
2
1
2
21
2
1
+−−++=
−+−+−++=
αααααα
αααααx
−
−
−
=⇒
=
=
⇒
=−+⇒=
∂
=−+⇒=
∂
11
16
11
4
11
8
11
4
11
16
11
4
08250
2
04230
2
2
1
12
2
21
1 x
x
x
α
α
αα
α
αα
α
has the smallest Euclidean
norm ,
3.10 find the solution in problem 3.8 that has the smallest Euclidean norm 3.8 中 范
小 解,
−
+
−
=
0
1
2
1
1
0
0
1
αx for any real α
the Euclidean norm of x is
−
−
=⇒=⇒=−⇒=
∂
∂
+−=++−+−=
1
6
1
6
3
6
5
6
1
02120
2261)2()1( 2222
x
x
x
αα
α
ααααα
has the
smallest Euclidean norm
3.11 consider the equation ]1[]2[]1[]0[]0[}{ 21 −+−++++= −− nubnubAubAubAxAnx nnn
where A is an nn× matrix and b is an 1×n column vector ,under what conditions on A and b will there
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exist ]1[],1[],[ −nuuou to meet the equation for any ]0[],[ xandnx ?
令 A nn× , b 1×n 列向 , 在 A 和b 足什么 件 ,存在 ]1[],1[],[ −nuuou ,对所
]0[],[ xandnx ,它们都 足 ]1[]2[]1[]0[]0[}{ 21 −+−++++= −− nubnubAubAubAxAnx nnn ,
write the equation in this form
−
−
== −
]0[
]2[
]1[
],[]0[}{ 1
u
nu
nu
bAbAbxAnx nn
where ],[ 1bAbAb n− is an nn× matrix and ]0[}{ xAnx n− is an 1×n column vector , from the equation we
can see , ]1[],1[],[ −nuuou exist to meet the equation for any ]0[],[ xandnx ,if and only if
nbAbAb n =− ],[ 1r under this condition , there will exist ]1[],1[],[ −nuuou to meet the equation for any
]0[],[ xandnx .
3.12 given
=
=
=
1
3
2
1
1
0
0
1000
0200
0120
0012
bbA what are the representations of A with respect to
the basis { }bAbAbAb 32 and the basis { }bAbAbAb 32 , respectively? 给定
=
=
=
1
3
2
1
1
0
0
1000
0200
0120
0012
bbA 请 A 关于{ }bAbAbAb 32 和基{ }bAbAbAb 32 表
分别 什么?
=⋅=
=⋅=
=
1
16
32
24
,
1
4
4
1
,
1
2
1
0
232 bAAbAbAAbAbA
we have [ ] [ ]
−
−
==∴
+−+−=
7100
18010
20001
8000
718208
3232
324
bAbAbAbbAbAbAbA
bAbAbAbbA
thus the representation of A
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with respect to the basis { }bAbAbAb 32 is
−
−
=
7100
18010
20001
8000
A
bAbAbAbbA
bAbAbAbA
324
432
718208
1
48
128
152
,
1
24
52
50
,
1
12
20
15
,
1
6
7
4
+−+−=
=
=
=
=
[ ] [ ]
−
−
==∴
7100
18010
20001
8000
3232 bAbAbAbbAbAbAbA thus the representation of A with
respect to the basis { }bAbAbAb 32 is
−
−
=
7100
18010
20001
8000
A
3.13 find Jordan-form representations of the following matrices
写出下列 jordan 型表 :
.
20250
16200
340
.
200
010
101
.
342
100
010
,
300
020
1041
4321
−−
=
−
=
−−−
=
= AAAA
the characteristic polynomial of 1A is )3)(2)(1()det( 11 −−−=−=∆ λλλλ AI thus the eigenvelues of 1A are
1 ,2 , 3 , they are all distinct . so the Jordan-form representation of 1A will be diagonal .
the eigenvectors associated with 3,2,1 === λλλ ,respectively can be any nonzero solution of
[ ]
[ ]
[ ]′=⇒=
′=⇒=
′=⇒=
1053
0142
001
3331
2221
1111
qqqA
qqqA
qqqA
thus the jordan-form representation of 1A with respect to
{ }321 qqq is
=
300
020
001
ˆ
1A
the characteristic polynomial of 2A is
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2
23
22 )1)(1)(1(243()det()( AiiAI −++++=+++=−=∆ λλλλλλλλ has
eigenvalues jandj −−+−− 11,1 the eigenvectors associated with
jandj −−+−− 11,1 are , respectively
[ ] [ ] [ ]′−−′−−′− jjandjj 211211,111 the we have
QAQ
j
jAand
jj
jjQ 2
1
2
100
010
001
ˆ
220
111
111
−=
−−
+−
−
=
−
−−+−−=
the characteristic polynomial of 3A is )2()1()det()(
2
33 −−=−=∆ λλλλ AI theus the
eigenvalues of 3A are 1 ,1 and 2 , the eigenvalue 1 has multiplicity 2 , and nullity213)(3)( 33 =−=−−=− IArankIA the 3A has two tinearly independent eigenvectors
associated with 1 ,
[ ] [ ]
[ ]′−=⇒=−
′=′=⇒=−
1010)(
0100010)(
333
213
qqIA
qqqIA
thus we have
QAQAandQ 3
1
3
200
010
001
ˆ
100
011
101
−=
=
−
−=
the characteristic polynomial of 4A is
3
44 )det()( λλλ =−=∆ AI clearly 4A has lnly one
distinct eigenvalue 0 with multiplicity 3 , Nullity( 4A -0I)=3-2=1 , thus 4A has only one
independent eigenvector associated with 0 , we can compute the generalized , eigenvectors
of 4A from equations below
[ ]
[ ]
[ ]′−=⇒=
′−=⇒=
′=⇒=
430
540
0010
3231
2121
111
vvvA
vvvA
vvA
, then the representation of 4A
with respect to the basis { }321 vvv is
−
−==
= −
450
340
001
000
100
010
ˆ
4
1
4 QwhereQAQA
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3.14 consider the companion-form matrix
−−−−
=
0100
0010
0001
4321 αααα
A show that its
characterisic polynomial is given by 43
3
2
3
1
4)( αλαλαλαλλ ++++=∆
show also that if iλ is an eigenvalue of A or a solution of 0)( =∆ λ then
[ ]133 ′iii λλλ is an eigenvector of A associated with iλ
证 友 A 征多 式 43
3
2
3
1
4)( αλαλαλαλλ ++++=∆ 并且,如 iλ iλ 一
个 征值, 0)( =∆ λ 一个解, 那么向 [ ]133 ′iii λλλ A 关于 iλ 一个 征向
proof:
4322
3
1
4
43
2
3
1
4
432
1
4321
2
1
det
1
0
det
10
01det
10
01
00
det)(
100
010
001
det)det()(
αλαλαλαλ
λ
αα
λ
λ
αλαλ
λ
λ
ααα
λ
λ
λ
αλ
λ
λ
λ
ααααλ
λλ
++++=
−
+
−
++=
−
−+
−
−+=
−
−
−
+
=−=∆ AI
if iλ is an eigenvalue of A , that is to say , 0)( =∆ iλ then we have
=
=
−−−−
=
−−−−
1110100
0010
0001 2
3
2
3
4
2
43
2
2
3
1
2
3
4321
i
i
i
I
i
i
i
i
i
i
iii
i
i
i
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
αλαλαλα
λ
λ
λαααα
that is to say [ ]133 ′iii λλλ is an eigenvetor oa A associated with iλ
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3.15 show that the vandermonde determinant
1111
4321
2
4
2
3
2
2
2
1
3
4
3
3
3
2
3
1
λλλλ
λλλλ
λλλλ
equals
)(41 ijji λλ −Π ≤<≤ , thus we conclude that the matrix is nonsingular or equivalently , the
eigenvectors are linearly independent if all eigenvalues are distinct ,证 vandermonde 行列式
1111
4321
2
4
2
3
2
2
2
1
3
4
3
3
3
2
3
1
λλλλ
λλλλ
λλλλ
为 )(41 ijji λλ −Π ≤<≤ , 因 如 所 征值都互不
同则该 奇异, 或者 价地说, 所 征向 性 关,
proof:
41
241423132412
23132414424142313313
12
24142313
2414423133
141312
24142313
2414423133
141312
144133122
14
2
413
2
312
2
2
141312
144133122
14
2
413
2
312
2
2
4321
2
4
2
3
2
2
2
1
3
4
3
3
3
2
3
1
))()()()()((
)])()()(())()()(()[(
)(
))(())((
))(())((
det
))(())((0
))(())((0
det
)()()(
)()()(
det
1111
0
)()()(0
)()()(0
det
1111
det
≤<≤Π=
−−−−−−=
−−−−−−−−−−−=
−⋅
−−−−
−−−−
−=
−−−
−−−−
−−−−
−=
−−−
−−−
−−−
=
−−−
−−−
−−−
=
ji
λλλλλλλλλλλλ
λλλλλλλλλλλλλλλλλλλλ
λλ
λλλλλλλλ
λλλλλλλλλλ
λλλλλλ
λλλλλλλλ
λλλλλλλλλλ
λλλλλλ
λλλλλλλλλ
λλλλλλλλλ
λλλλλλ
λλλλλλλλλ
λλλλλλλλλ
λλλλ
λλλλ
λλλλ
let a, b, c and d be the eigenvalues of a matrix A , and they are distinct Assuming the matrix is
singular , that is abcd=0 , let a=0 , then we have
))()((
111
detdet
1111
0
0
0
det
222
222
333
222
333
bcbdcdbcddcb
dcb
bcd
dcb
dcb
dcb
dcb
dcb
dcb
−−−−=
⋅−=
=
and from vandermonde determinant
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))(())()()((
1111
0
0
0
det
1111
det
222
333
2222
3333
bdcdbcdabacbdcd
dcb
dcb
dcb
dcba
dcba
dcba
−−=−−−−=
=
so we can see
−=
1111
0
0
0
det
1111
0
0
0
det
222
333
222
333
dcb
dcb
dcb
dcb
dcb
dcb
,
that is to say dandcbabdcdbcd ,,,0))(( ⇒=−− are not distinet
this implies the assumption is not true , that is , the matrix is nonsingular let
i
q be the
eigenvectors of A ,
44332211
qaqAqaqAqaqAqaqA ====
( )
tindependenlinenrlyq
qand
q
so
ddd
ddc
bbb
aaa
qqqq
qdqcqbqa
qdqcqbqa
qdqcqbqa
qqqq
ii
ni
nii ⇒=
≠
=
=
⇒
=+++
=+++
=+++
=+++
×
×
×
0
0
0
0
1
1
1
1
0
0
0
0
1
1
44
32
32
32
32
44332211
44
3
33
3
22
3
11
3
44
2
33
2
22
2
11
2
44332211
44332211
α
α
αααα
αααα
αααα
αααα
αααα
3.16 show that the companion-form matrix in problem 3.14 is nonsingular if and only if
04 ≠α , under this assumption , show that its inverse equals
−−−−
=−
4
3
4
2
4
1
4
1
1
1000
0100
0010
α
α
α
α
α
α
α
A 证 3.14 中 友 奇异当且仅当
04 ≠α , 且 逆为
−−−−
=−
4
3
4
2
4
1
4
1
1
1000
0100
0010
α
α
α
α
α
α
α
A
proof: as we know , the chacteristic polynomial is
43
2
2
3
1
4)det()( αλαλαλαλλλ ++++=−=∆ AI so let 0=λ , we have
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4
4 )det()det()1()det( α==−= AAA
A is nonsingular if and only if
04 ≠α
−−−−
=∴
=
⋅
−−−−
⋅
−−−−
=
⋅
−−−−
⋅
−−−−
−
4
3
4
2
4
1
4
1
4
4
3
4
2
4
1
4
4321
4
4321
4
3
4
2
4
1
4
1
1000
0100
0010
1000
0100
0010
0001
1
1000
0100
0010
0100
0010
0001
1000
0100
0010
0001
0100
0010
0001
1
1000
0100
0010
α
α
α
α
α
α
α
α
α
α
α
α
α
α
αααα
αααα
α
α
α
α
α
α
α
A
I
I
3.17 consider
=
λ
λλ
λλλ
00
0
2/2
tT
tTT
A with 0≠λ and T>0 show that [ ]′100 is an
generalized eigenvector of grade 3 and the three columns of
=
000
00
02/222
tT
TT
Q λ
λλ
constitute a chain of generalized eigenvectors of length 3 , venty
=−
λ
λ
λ
00
10
01
1 tAQQ
A 中 0≠λ ,T>0 , 证 [ ]′100 3 广义 征向 , 并且 Q 3 列 成 度
3 广义 征向 ,验证
=−
λ
λ
λ
00
10
01
1 tAQQ
Proof : clearly A has only one distinct eigenvalue λ with multiplicity 3
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0
1
0
0
000
000
000
1
0
0
000
00
2/0
000
000
00
1
0
0
)(
0
0
0
1
0
0
000
000
00
1
0
0
)(
222
3
2222
2
=
=
=
=−
≠
=
=
=−
T
TTT
IA
TT
IA
λ
λλλ
λ
λλ
λ
these two equation imply that [ ]′100 is a generalized eigenvctor of grade 3 ,
and
=
=
=−
=
=
=−
0
0
0000
00
2/0
0
)(
01
0
0
000
00
2/0
1
0
0
)(
2222222
222
T
T
T
T
TT
T
T
IA
T
T
T
TT
IA
λ
λ
λ
λ
λλ
λ
λ
λ
λ
λ
λ
λλ
λ
that is the three columns of Q consititute a chain of generalized eigenvectors of length 3
=
=
=
=
=
−
λ
λ
λ
λ
λλ
λλλ
λ
λ
λ
λ
λλ
λ
λ
λ
λ
λλ
λλλ
λ
λλ
λ
λλ
λλλ
00
10
01
00
0
2/2/3
00
10
01
100
00
02/
00
10
01
00
0
2/2/3
100
00
02/
00
0
2/
1
2
22223222
222232222
AQQ
TT
TTT
T
TT
Q
TT
TTT
T
TT
T
TT
AQ
3.18 Find the characteristic polynomials and the minimal polynomials of the following matrices
下列 征多 式和 小多 式,
)(
000
000
000
000
)(
000
000
000
001
)(
000
000
010
001
)(
000
000
010
001
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
1
dcba
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
λ
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)()()()()(
)()()()()(
)()()()()(
)()()()()()()(
14
4
14
2
13
4
13
3
12
4
12
2
3
112
3
11
λλλλλλ
λλλλλλ
λλλλλλ
λλλλλλλλλλ
−=Ψ−=∆
−=Ψ−=∆
−=Ψ−=∆
−−=Ψ−−=∆
d
c
b
a
3.19 show that if λ is an eigenvalue of A with eigenvector x then )(λf is an eigenvalue
of )(Af with the same eigenvector x
证 如 λ A 关于λ 征向 ,那么 )(λf )(Af 征值, x )(Af 关于
)(λf 征向 ,
proof let A be an nn× matrix , use theorem 3.5 for any function )(xf we can define
1
110)(
−
−+++=
n
n xxxh bbb which equals )(xf on the spectrum of A
if λ is an eigenvalue of A , then we have )()( λλ hf = and
)()( AhAf =
xfxh
xxx
xAAIxAhxAf
xxAxxAxxAxAxxAf
n
n
n
n
kk
)()(
)()()(
,,)(
1
110
1
110
3322
λλ
λbλbb
bbb
λλλλλλ
==
+++=
+++==∴
====⇒
−
−
−
−
which implies that )(λf is an eigenvalue of )(Af with the same eigenvector x
3.20 show that an nn× matrix has the property 0=kA for mk ≥ if and only if A has
eigenvalues 0 with multiplicity n and index m of less , such a matrix is called a nilpotent matrix
证 nn× 在 0=kA 当且仅当A n 重 0 征值指 不大于m ,这 被 为
归 ,
proof : if A has eigenvalues 0 with multiplicity n and index M or less then the Jordan-form
representation of A is
=
3
2
1
ˆ
J
J
J
A where
mnma
nn
J
i
i
l
i
i
nn
i
ii
≤×
=
= ∑=
×
1
0
10
10
from the nilpotent property , we have i
k
i nkforJ ≥= 0 so if
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,i
i
nmamk ×≥≥ liallJ ki ,,2,10 == and 0ˆ
3
2
1
=
=
k
k
k
J
J
J
A
then )0)ˆ(0)((0 === AfifonlyandifAfAk
If ,0 mkforAk ≥= then ,0ˆ mkforAk ≥= where
=
3
2
1
ˆ
J
J
J
A ,
nn
J
l
i
i
nni
i
i
i
ii
=
= ∑
=
×
1
1
1
λ
λ
λ
So we have
liformnand
li
nnk
k
k
J
ii
k
i
k
i
nk
i
ii
k
i
k
i
i
k
i
,2,1,0
,2,10
)!1)(1(
! 11
=≤=∴
==
−+−
=
+−−
λ
λ
λ
λλλ
which implies that A has only one distinct eigenvalue o with multiplicity n and index m or less ,
3.21 given
=
100
100
011
A , find AteandAA 10310 , A 函 AteandAA 10310 , ,
the characteristic polynomial of A is
2
2 )1()det()( −=−=∆ λλλλ AI
let
2
210)( λbλbbλ ++=h on the spectrum of A , we have
21
10
210
10
0
10
2110)1()1(
1)1()1(
0)0()0(
bb
bbb
b
+=⋅′=′
++==
==
hf
hf
hf
the we have 98,0 220 =−== bbb
=
+
−=
+−=
100
100
911
100
100
101
9
100
100
011
8
98 210 AAA
the compute
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102
101
0
21103
1
0
2
1
0
21
103
210
103
0
103103
=
−=
=
⇒
+=⋅
++=
=
b
b
b
bb
bbb
bA
=
+
−=
+−=
100
100
10211
100
100
011
102
100
100
011
101
102101 2103 AAA
to compute
Ate :
1
22
1
2 2
1
0
21
210
0
0
+−=
−−=
=
⇒
+=
++=
=
tt
tt
t
t
ete
tee
te
e
e
b
b
b
bb
bbb
b
−
+−−
=
+−+
−−+
=
++=
t
t
tttt
tttt
At
e
e
eteee
eeee
AAIAe
00
110
11
100
100
111
)12(
100
100
011
)22(
100
010
001
2
210 bbb
3.22 use two different methods to compute
Ate for A1 and A4 in problem 3.13
两 计 3.13 中A1和A4 函
tAAt ee 2,
method 1 : the Jordan-form representation of A1 with respect to the basis
[ ] [ ] [ ]{ }′′′ 105014001 is { }3,2,1ˆ1 diagA =
−−
=
=
=
==∴
−
−
t
t
ttttt
t
t
ttt
t
t
t
tAtA
e
e
eeeee
e
e
eee
Cd
Cc
e
e
e
Cb
QwhereQQeeCa
3
2
33
3
2
32
1
3
2
1ˆ
00
00
)(5)(4
100
010
541
00
00
54
100
010
541
00
00
00
100
010
541
100
010
541
,11
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+−−
+
++
=
−
=
=∴
−
−==
=
−
120250
161200
232/541
100
10
2/1
450
340
001
450
340
001
000
100
010
ˆ
22
2
1
4
44
4
tt
tt
tttt
t
tt
QQee
QwhereQQeA
tAtA
tA
method 2: the characteristic polynomial of 1A is ).3)(2)(1()( −−−=∆ λλλλ let
2
210)( λbλbbλ ++=h on the spectrum of 1A , we have
)2(
2
1
2
3
4
2
5
33
93)3()3(
32)2()2(
)1()1(
32
3
32
2
33
0
2210
3
210
2
10
ttt
ttt
ttt
t
t
t
eee
eee
eee
ehf
ehf
ehf
+−=
−++−=
+−=
⇒
++==′
++==
++==
b
b
b
bbb
bbb
bbb
−−
=
+−+
−+−+
+−
+−
+−
=
++=
t
t
ttttt
ttt
ttt
ttt
ttt
ttt
tA
e
e
eeeee
eee
eee
eee
eee
eee
AAIe
3
2
32
32
32
32
32
32
2
12210
00
00
)(5)(4
900
040
40121
)2(
2
1
300
020
1041
)
2
3
4
2
5
(
3300
0330
0033
1 bbb
the characteristic polynomial of 4A is
3)( λλ =∆ , let 2210)( λbλbbλ ++=h on the
spectrum of 4A ,we have
2
2
1
0
2:)0()0(
:)0()0(
1:)0()0(
b
b
b
=′′=′′
=′=′
==
thf
thf
hf
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thus
+−−
+
++
=
+
−−
+
+=
++=
120250
161200
232/541
000
000
450
2/
20250
16200
2340
22
2
2
2
41410
4
tt
tt
tttt
t
tt
tI
AAIe
tA bbb
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3.23 Show that functions of the same matrix ; that is )()()()( AfAgAgAf = consequently
we have AeAe AtAt = 证 同一 函 具 可交换性，即 )()()()( AfAgAgAf = 因
AeAe AtAt = 成
proof: let
1
110
1
110
)(
)(
−
−
−
−
+++=
+++=
n
n
n
n
AAIAf
AAIAg
ααα
bbb
( n is the order of A)
k
ik
k
nki
i
n
nk
k
ik
k
i
i
n
k
n
nn
n
in
n
i
i
n
in
n
i
i
AA
AAAAIAgAf
)()(
)()()(
1
22
0
1
0
22
11
1
1
1
1
0
011000
−
+−=
−
=
−
=
−
=
−
−−−
=
−
−−
−
=
∑∑∑∑
∑∑
+=
+++++++=
bαbα
bαbαbαbαbαbα
)()()()()()(
1
22
0
1
0
AgAfAAAgAf kik
k
nki
i
n
nk
k
ik
k
i
i
n
k
=+= −
+−=
−
=
−
=
−
=
∑∑∑∑ bαbα
let
AteAgAAf == )(,)( then we have AeAe AtAt =
3.24 let
=
3
2
1
00
00
00
λ
λ
λ
C , find a B such that CeB = show that of 0=iλ for some
I ,then B does not exist
let
=
3
2
1
00
00
00
λ
λ
λ
C , find a B such that CeB = Is it true that ,for any nonsingular c ,there
exists a matrix B such that CeB = ?
令
=
3
2
1
00
00
00
λ
λ
λ
C 证 若 0111 =⋅⋅ λλλ ,则不存在 B 使 Ce
B =
若
=
3
2
1
00
00
00
λ
λ
λ
C , 否对任意 奇异 C 都存在 B 使, CeB = ?
Let λλ λ == ef ln)( so BeBf B == ln)(
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==
3
2
1
ln00
0ln0
00ln
ln
λ
λ
λ
CB where 3,2,1,01 => iλ , if 0=λ for some i ,
1lnλ does not exist
for
=
λ
λ
λ
00
00
01
C we have
=
′
==
λ
λ
λλ
λ
λ
λλ
ln00
0ln0
01ln
ln00
0ln0
0nlln
lnCB ,
where 0,0 ≤> λλ if then λln does mot exist , so B does mot exist , we can conclude
that , it is mot true that , for any nonsingular C THERE EXISTS a B such that cek =
3.25 let )(
)(
1
)( AsIAdj
s
AsI −
∆
=− and let m(s) be the monic greatest common divisor of
all entries of Adj(Si-A) , Verify for the matrix 2A in problem 3.13 that the minimal polynomial
of A equals )()( sms∆
令, )(
)(
1
)( AsIAdj
s
AsI −
∆
=− , 并且令 m(s) Adj(Si-A) 所 元 一 大公因子,
利 3.13 中 2A 验证 A 小多 式为 )()( sms∆
verification :
1)(
)1(00
0)2)(1(0
)1(0)2)(1(
)(,
200
010
101
)2(\)1()()2()1()(
200
010
001
ˆ
200
010
101
2
2
33
−=
−
−−
−−−−
=−
−
−
−
=−
−−=−−=∆
=
−
=
ssmso
s
ss
sss
AsIAdj
s
S
S
AsI
ssssss
AA
j
we can easily obtain that )()()( smss ∆=j
3.26 Define [ ]1221101
)(
1
)( −−
−−− ++++
∆
=− nn
nn RsRsRsR
s
AsI where
in
nn RandsssAsIs ααα ++++=−=∆ −− 22
1
1
2:)det()( are constant matrices theis
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definition is valid because the degree in s of the adjoint of (sI-A) is at most n-1 , verify
IAR
n
ARtr
IAAAIARR
n
ARtr
IAAIARR
ARtr
IAIARR
ARtr
IR
ARtr
nnn
nn
nn
nnnn
αα
ααααα
αααα
ααα
α
+=−=
++++=+=
−
−=
++=+=−=
+=+=−=
=−=
−
−−
−−
−−−−
1
1
12
2
1
1
121
1
1
21
2
212
1
3
1101
1
2
0
0
1
0
)(
1
)(
2
)(
2
)(
1
)(
where tr stands for the trase of a matrix and is defined as the sum of all its diagonal entries this
process of computing ii Randα is called the leverrier algorithm
定义 ][
)(
1
:)( 12
2
1
1
0
1
−−
−−− ++++
∆
=− nn
nn RSRSRSR
s
ASI 其中 )(s∆ A 征多
式 in
nnn RandsssAsIs ααα ++++=−=∆ −− 22
1
1:)det()( 常 ,这 定义
, 因为 SI-A 伴 中 S 不超过 N-1 验证
IAR
n
ARtr
IAAAIARR
n
ARtr
IAAIARR
ARtr
IAIARR
ARtr
IR
ARtr
nnn
nn
nn
nnnn
αα
ααααα
αααα
ααα
α
+=−=
++++=+=
−
−=
++=+=−=
+=+=−=
=−=
−
−−
−−
−−−−
1
1
12
2
1
1
121
1
1
21
2
212
1
3
1101
1
2
0
0
1
0
)(
1
)(
2
)(
2
)(
1
)(
其中 迹 tr 定义为其对角元 之和, 这 计 iα 和 iR 式被 为 leverrier .
verification:
implieswhich
SI
SSSSI
ARSARRSARRSARRSR
RSRSRSRASI
nn
nnn
nnn
nnn
nn
nn
)(
)(
)()()(
])[(
1
2
2
1
1
121
2
12
1
010
12
2
1
1
0
∆=
+++++=
+−+−+−+=
++++−
−
−−
−−−
−−
−−
−−
αααα
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][
)(
1
:)( 12
2
1
1
0
1
−−
−−− ++++
∆
=− nn
nn RSRSRSR
s
ASI ’
where n
nnn ssss ααα ++++=∆ −− 22
1
1)(
3.27 use problem 3.26 to prove the cayley-hamilton theorem
利 3.26 证 cayley-hamilton 定
proof:
IAR
IARR
IARR
IARR
IR
nn
nnn
α
α
α
α
=−
=−
=−
=−
=
−
−−−
1
121
212
101
0
multiplying ith equation by
1+−inA yields ( ni 2,1= )
IAR
ARAAR
ARARA
ARARA
ARA
nn
nnn
nnn
nnn
nn
α
α
α
α
=−
=−
=−
=−
=
−
−−−
−−−
−−
1
12
2
1
2
21
1
2
2
1
101
1
0
then we can see
012
2
101
1
0
1
2
2
1
1
=−−++−++=
+++++
−−−
−
−
−−
nnn
nnn
nn
nnn
ARRAARRARARA
IAAAA
αααα
that is
0)( =∆ A
3.28 use problem 3.26 to show
[ ]IssAssAsA
s
AsI
n
nnnnn )()()(
)(
1
)(
1
2
1
13
21
22
1
1
1
−
−−−−−
−
++++++++++
∆
=
−
ααααα
利 3.26 证 上式,
Proof:
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[ ]IssAssAsA
s
IAAASAAA
SIAASIAS
s
RSRSRSR
s
ASI
n
nnnnn
nn
nn
nn
nn
nnn
nn
nn
)()()(
)(
1
])(
)()([
)(
1
][
)(
1
:)(
1
2
1
13
21
22
1
1
12
2
1
1
43
3
1
2
3
21
22
1
1
12
2
1
1
0
1
−
−−−−−
−−
−−
−−
−−
−−−
−−
−−−
++++++++++
∆
=
+++++++++
++++++
∆
=
++++
∆
=−
ααααα
αααααα
ααα
another : let 10 =α
[ ]IssAssAsA
s
IAAASAAA
SIAASIAS
ASAS
s
AS
s
SR
s
RSRSRSR
s
ASI
n
nnnnn
nn
nn
nn
nn
nnn
n
i
In
i
N
i
In
i
n
ii
n
i
L
li
n
i
inin
i
nn
nn
)()()(
)(
1
])(
)()(
)(
1
)(
1
)(
1
][
)(
1
:)(
1
2
1
13
21
22
1
1
12
2
1
1
43
3
1
2
3
21
22
1
1
1
0
1
1
0
01
1 1
0
1
0
11
12
2
1
1
0
1
−
−−−−−
−−
−−
−−
−−
−−−
−
=
−−
−
=
−−
−
=
−
=
−
−
=
−−−−
−−
−−−
++++++++++
∆
=
+++++++++
++++++
+++
∆
=
∆
=
∆
=
++++
∆
=−
∑∑
∑ ∑∑
ααααα
αααααα
ααα
αα
α
3.29 let all eigenvalues of A be distinct and let iq be a right eigenvector of A associated with
iλ that is iIi qqA λ= define [ ]nqqqQ 21= and define
== −
np
p
p
QP
2
1
1 :; ,
where ip is the ith row of P , show that ip is a left eigenvector of A associated with iλ , that
is iii pAp λ= 如 A 所 征值互不 同 , iq 关于 iλ 一个右 征向 ,即
iIi qqA λ= ,定义 [ ]nqqqQ 21= 并且
== −
np
p
p
QP
2
1
1 :;
其中 ip P I 行, 证 ip A 关于 iλ 一个左 征向 ,即 iii pAp λ=
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Proof: all eigenvalues of A are distinct , and iq is a right eigenvector of A associated with iλ ,
and [ ]nqqqQ 21= so we know that 11
3
2
1
ˆ −− ==
= PAPAQQA
λ
λ
λ
PAPA ˆ=∴ That is
=
⇒
=
nnnn
n
n p
p
p
Ap
Ap
Ap
p
p
p
A
p
p
p
λ
λ
λ
λ
λ
λ
22
11
2
1
2
1
2
1
2
1
: so iii pAp λ= , that is , ip is a
left eigenvector of A associated with iλ
3.30 show that if alleigenvalues of A are distinct , then
1)( −− ASI can be expressed as
∑ −=−
−
ii
i
pq
s
ASI
λ
1
)( 1 where iq and ip are right and left eigenvectors of A associated
with iλ 证 若 A 所 征 值 互 不 同 , 则
1)( −− ASI 可 以 表 为
∑ −=−
−
ii
i
pq
s
ASI
λ
1
)( 1 其中 iq 和 ip A 关于 iλ 右 征值和左 征值,
Proof: if all eigenvalues of A are distinct , let iq be a right eigenvector of A associated with iλ ,
then [ ]nqqqQ 21= is nonsingular , and
=−
np
p
p
Q
2
1
1: where is a left eigenvector of
A associated with iλ ,
∑ ∑
∑∑
∑
=−
−
=
−
−
=−
−
=
−
−
iiiii
i
iiiii
i
iiii
i
ii
i
pqpqs
s
pqpqs
s
Apqpqs
s
ASIpq
s
))((
1
)(
1
)(
1
)(
1
λ
λ
λ
λλ
λ
That is ∑ −=−
−
ii
i
pq
s
ASI
λ
1
)( 1
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3.31 find the M to meet the lyapunov equation in (3.59) with
==
−−
=
3
3
3
22
10
CBA what are the eigenvalues of the lyapunov equation ? is the
lyapunov equation singular ? is the solution unique ?
已 A,B,C, M 使之 足(3.59) lyapunov 征值, 该 否奇异>解 否唯一?
=⇒=
−
∴
=+
3
0
12
13
MCM
CMBAM
3)det()()1)(1()det()( +=−=∆++−+=−=∆ λλλλλλλ BIjjAI BA
The eigenvalues of the Lyapunov equation are
jjjj −=+−−=+=++−= 231)231 21 ηη
The lyapunov equation is nonsingular M satisfying the equation
3.32 repeat problem 3.31 for
−
=
==
−−
=
3
3
3
3
1
21
10
21 CCBA with two
different C , 给出 A, B, C, 重复 3.31 ,
α
α
α
anyforMCM
solutionNoCM
CMBAM
−
=⇒=
−−
⇒=
−−
∴
=+
311
11
11
11
2
1
1)()1()( 2 −=∆+=∆ λλλλ BA
the eigenvalues of the lyapunov equation are 01121 =+−==ηη the lyapunov equation is
singular because it has zero eigenvalue if C lies in the range space of the lyapunov equation , then
solution exist and are not unique
,
3.33 check to see if the following matrices are positive definite or senidefinite 定下列
否 定或者 半定,
−
−
332313
322212
312111
201
000
100
202
013
232
aaaaaa
aaaaaa
aaaaaa
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1.
∴<−=−=
202
013
232
0792
13
32
det is not positive definite , nor is pesitive
semidefinite
2. )21)(21(
201
00
10
det −−+−=
−
−
λλλ
λ
λ
λ
it has a negative eigenvalue 21− ,
so the second matrix is not positive definite , neither is positive demidefinte ,,
3 the third matrix ‘s prindipal minors ,0,0,0 332211 ≥≥≥ aaaaaa `
0det0det0det0det
332313
322212
312111
3323
3222
3313
3111
2212
2111 =
=
=
=
aaaaaa
aaaaaa
aaaaaa
aaaa
aaaa
aaaa
aaaa
aaaa
aaaa
that is all the principal minors of the thire matrix are zero or positive , so the matrix is positive
semidefinite ,
3.34 compute the singular values of the following matrices 计 下列 奇值,
−
−
=
−
−
−
−
−
−
−
52
22
01
10
21
012
101
42
21
012
101
)6)(1(674)5)(2()( 2 −−=+−=−−−=∆ λλλλλλλ
the eigenvalues of
−
−
−
−
01
10
21
012
101
are 6 and 1 , thus the singular values of
−
−
012
101
are 6 and 1 ,
=
−
−
206
65
42
21
42
21
)41
2
3
2
25
)(41
2
3
2
25
(36)20)(5()( −−+−=−−−=∆ λλλλλ the eigenvalues of
−
−
42
21
42
21
are 41
2
3
2
25
41
2
3
2
25
−+ and , thus the singular values of
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−
42
21
are 70.1)41
2
3
2
25
(70.4)41
2
3
2
25
( 2
1
2
1
=−=+ and
3.35 if A is symmetric , what is the ralatimship between its eigenvalues and singular values ? 对
A 征值与奇异值之 什么关 ?
If A is symmetric , then
2AAAAA =′=′ LET iq be an eigenvector of A associated with
eigenvaue iλ that is , ),3,1(, niqqA iii == λ thus we have
),3,1(
22 niqqAqAqA iiiiiii ==== λλλ
Which implies
2
iλ is the eigenvalue of niforA ,2,1
2 = , (n is the order of A )
So the singular values of A are || iλ nifor ,2,1= where iλ is the eigenvalue of A
3.36 show [ ] m
n
m
mn
n
n babbb
a
a
a
I ∑
=
+=
+
1
21
2
1
1det
证 上式成
let [ ]n
n
bbbB
a
a
a
A
21
2
1
=
= A is 1×n and B is n×1
use (3.64) we can readily obtain
[ ] [ ]
∑
≤
+=
+=
+
n
m
mm
n
nn
n
n
ba
a
a
a
bbbIbbb
a
a
a
I
1
2
1
21121
2
1
1
detdet
3.37 show (3.65) 证 (3.65)
proof: let
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−
−
=
=
=
n
m
n
m
n
m
IsB
AIs
P
IsB
Is
Q
Is
AIs
N
0
0
then we have
−
−
=
−
=
BAsI
AssI
QP
sIBs
ABsI
NP
n
m
n
m
0
0
because
PsPQQP
PsPNNP
detdetdet)det(
detdetdet)det(
==
==
we have det(NP)=det(QP)
And
)det()det()det()det(
)det()det()det()det(
BAsISBAsIsIQP
ABsISsIABsINP
n
m
nm
m
n
nm
−=−=
−=−=
3.38 Consider yxA = , where A is nm× and has rank m is yAAA ′′ −1)( a solution ? if not ,
under what condition will it be a solution? Is yAAA 1)( −′′ a solution ?
nm× A 为 m , yAAA ′′ −1)( 不 yxA = 解? 如 不 , 那么在
什么 件下, 他才会成为该 解? yAAA 1)( −′′ 不 解?
A is nm× and has rank m so we know that nm ≤ , and AA′ is a square matrix of mm×
rankA=m , rank( nmrankAAA ≤=≤′ ) ,
So if nm ≠ ,then rank( AA′ )<n , 1)( −′AA does mot exist m and yAAA ′′ −1)( isn’t a
solution if yxA =
If m=n , and rankA=m , so A is nonsingular , then we have rank( AA′ )=rank(A)=m , and
A yAAA ′′ −1)( =A yyAAA =′′ −− 11 )( that is yAAA ′′ −1)( is a solution ,
RankA=M
∴ Rank( AA′ )=m AA′ is monsingular and 1)( −′AA exists , so we have
yyAAAA =′′ −1)( , that is , yAAA 1)( −′′ is a solution of yxA = ,
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PROBLEMS OF CHAPTER 4
4.1 An oscillation can be generated by 一个振荡器可 下式 述：
XX
−
=
01
10
试证其解为：
Show that its solution is )0(
cossin
sincos
)( X
tt
tt
tX
−
=
Proof: )0()0()(
01
10
XeXetX
t
At
−== ,the eigenvalues of A are j,-j;
Let λbbλ 10)( +=h .If
teh λλ =)( ,then on the spectrum of A,then
tjtejjh
tjtejjh
jt
jt
sincos)(
sincos)(
10
10
−==−=−
+==+=
−bb
bb
then
t
t
sin
cos
1
0
=
=
b
b
so
−
=
−
+
=+=
tt
tt
ttAIAh
cossin
sincos
01
10
sin
10
01
cos)( 10 bb
)0(
cossin
sincos
)0()0()(
01
10
X
tt
tt
XeXetX
t
At
−
===
−
4.2 Use two different methods to find the unit-step response of 两 下 统 单位
跃响应：
UXX
+
−−
=
1
1
22
10
[ ]XY 32=
Answer: assuming the initial state is zero state.
method1:we use (3.20) to compute
−
+
++
=
+
−
=−
−
−s
s
sss
s
AsI
2
12
22
1
22
1
)(
2
1
1
then
tAt e
ttt
ttt
AsILe −−−
−−
+
=−=
sincossin2
sinsincos
))(( 11 and
)22(
5
)22(
5
)()()(
22
1
++
=
++
=−= −
sssss
s
sBUAsICsY
then tety t sin5)( −= for t>=0
method2:
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t
tt
tt
At
t
tA
t
tA
te
tete
tete
C
BeeCABdeC
dBueCty
−
−−
−−
−−
−
=
−−
−+
−−
=
−==
=
∫
∫
sin5
1sin3cos
1sin2cos
01
5.01
)(
)()(
01
0
)(
0
)(
τ
ττ
τ
τ
for t>=0
4.3 Discretize the state equation in Problem 4.2 for T=1 and T=π . 化习 4.3 中 态
，T 分别取 1 和π
Answer:
][][][
][][]1[
0
kDUkCXkY
kBUdekXekX
T
AAT
+=
+=+ ∫ αα
For T=1,use matlab:
[ab,bd]=c2d(a,b,1)
ab =0.5083 0.3096
-0.6191 -0.1108
bd =1.0471
-0.1821
[ ] ][32][
][
1821.0
0471.1
][
1108.06191.0
3096.05083.0
]1[
kXkY
kUkXkX
=
−
+
−−
=+
for T=π ,use matlab:
[ab,bd]=c2d(a,b,3.1415926)
ab =-0.0432 0.0000
-0.0000 -0.0432
bd =1.5648
-1.0432
[ ] ][32][
][
0432.1
5648.1
][
0432.00
00432.0
]1[
kXkY
kUkXkX
=
−
+
−
−
=+
4.4 Find the companion-form and modal-form equivalent equations of 统 价友形和 式
规范形。
UXX
+
−−
−
=
1
0
1
220
101
002
[ ]XY 011 −=
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Answer: use [ab ,bb,cb,db,p]=canon(a,b,c,d,’companion)
We get the companion form
ab = 0 0 -4
1 0 -6
0 1 -4
bb = 1
0
0
cb = 1 -4 8
db =0
p =1.0000 1.0000 0
0.5000 0.5000 -0.5000
0.2500 0 -0.2500
UXX
+
−
−
−
=
0
0
1
410
601
400
[ ]XY 841 −=
use use [ab ,bb,cb,db,p]=canon(a,b,c,d) we get the modal form
ab = -1 1 0
-1 -1 0
0 0 -2
bb = -3.4641
0
1.4142
cb = 0 -0.5774 0.7071
db = 0
p = -1.7321 -1.7321 -1.7321
0 1.7321 0
1.4142 0 0
UXX
−
+
−
−−
−
=
4142.1
0
4641.3
200
011
011
[ ]XY 7071.05774.00 −=
4.5 Find an equivalent state equation of the equation in Problem 4.4 so that all state variables have
their larest magnitudes roughly equal to the largest magnitude of the output.If all signals are
required to lie inside 10± volts and if the input is a step function with magnitude a,what is the
permissible largest a?找出习 4.4 中 价 态 使所 态变 大 几乎 于
输出 大值。如 所 信号 要在 10± 伏以内，输入为大小为 a 跃函 。 所
允许 大 a 值。
Answer: first we use matlab to find its unit-step response .we type
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a=[-2 0 0;1 0 1;0 -2 -2]; b=[1 0 1]';c=[1 -1 0]; d=0;
[y,x,t]=step(a,b,c,d)
plot(t,y,'.',t,x(:,1),'*',t,x(:,2),'-.',t,x(:,3),'--')
0 1 2 3 4 5 6
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
x2
x1
x3
y
so from the fig above.we know max(|y|)=0.55, max(|x1|=0.5;max(|x2|)=1.05,and max(|x3|)=0.52
for unit step input.Define ,,5.0, 332211 xxxxxx === then
UXX
+
−−
−
=
1
0
1
240
5.005.0
002
[ ]XY 021 −=
the largest permissible a is 10/0.55=18.2
4.6 Consider U
b
b
XX
+
=
2
1
0
0
λ
λ [ ]XccY 11=
where the overbar denotes complex conjugate.Verify that the equation can be transformed into
XCyuBXAX 1=+=
with [ ])Re(2)Re(2
1
010
11111 cbcbCBA λλλλλ
−=
=
+−
= ,by using the
transformation XQX = with
−
−
=
11
11
bb
bb
Q
λ
λ
。试验证以上变换成 。
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Answer: let XQX = with
−
−
=
11
11
bb
bb
Q
λ
λ
,we get
U
b
b
XQXQ
+
=
2
1
0
0
λ
λ
[ ] XQccY 11=
−
−
−
=−
11
11
11
1
)(
1
bb
bb
bb
Q
λλλλ
so
UBXAUX
U
bbbb
X
bb
bb
bb
bb
bb
U
b
b
bb
bb
bb
X
bb
bb
bb
bb
bb
U
b
b
QXQQX
+=
+
+−
=
−−
+
−
−
−
−
−
=
−
−
−
+
−
−
−
−
−
=
+
= −−
1
010
)(
0
)(
1
)(
1
)(
1
0
0
)(
1
0
0
111111
11
1
2
1
2
11
11
2
1
11
11
1111
11
11
11
11
2
111
λλλλ
λλλλλ
λ
λλ
λλ
λλ
λλλλλ
λ
λ
λ
λλλλ
λ
λ
[ ] [ ] [ ] XCXbcbccbcbX
bb
bb
ccXQccY 111111111
11
11
1111 =+−−=
−
−
== λλ
λ
λ
4.7 Verify that the Jordan-form equation
[ ]Xccccccy
U
b
b
b
b
b
b
XX
321321
3
2
1
3
2
1
1
1
0
1
1
=
+
=
λ
λ
λ
λ
λ
λ
can be transformed into
[ ]XCCCyu
B
B
B
X
A
IA
IA
X 3212
2
00
0
0
=
+
=
where iCandBA ,, are defined in Problem 4.6 and 2I is the unit matrix of order 2.
试验证变换成 ，其中 iCandBA ,, 习 4.6 中定义 形式。 2I 为二 单位 。
PROOF: Change the order of the state variables from [x1 x2 x3 x4 x5 x6]’ to [x1 x4 x2 x5 x3 x6]’
And then we get
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[ ] [ ]XCCCXccccccy
u
B
B
B
X
A
IA
IA
b
b
b
b
b
b
x
x
x
x
x
x
x
x
x
x
x
x
X
321332211
3
2
1
2
2
3
3
2
2
1
1
6
3
5
2
4
1
6
3
5
2
4
1
00
0
0
1
1
1
1
==
+
=
+
=
=
λ
λ
λ
λ
λ
λ
4.8 Are the two sets of state equations
[ ]XyUXX 011
0
1
1
100
220
212
−=
+
=
and
[ ]XyUXX 011
0
1
1
100
120
112
−=
+
−
=
equivalent?Are they zero-state equivalent?上 两 态 价吗？它们 态 价吗？
Answer:
[ ]
[ ]
2
2
2
1
1
1
)2(
1
0
1
1
)2(00
)2(2)1)(2(0
)1(2)1()1)(2(
)1()2(
011
0
1
1
100
220
212
011)()(ˆ
−
=
−
−−−
−−−−
−−
−
=
−
−−
−−−
−=−=
−
−
s
s
sss
ssss
ss
s
s
s
BAsICsG
[ ]
[ ]
2
2
2
1
1
2
)2(
1
0
1
1
)2(00
)2()1)(2(0
)1()1()1)(2(
)1()2(
011
0
1
1
100
120
112
011)()(ˆ
−
=
−
−+−
−++−
+−
−
=
+
−−
−−−
−=−=
−
−
s
s
sss
ssss
ss
s
s
s
BAsICsG
obviously, )(ˆ)(ˆ 21 sGsG = ,so they are zero-state equivalent but not equivalent.
4.9 Verify that the transfer matrix in (4.33)has the following realization:
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U
N
N
N
N
X
I
II
II
II
X
r
r
qr
qqr
qq
qq
+
−
−
−
−
=
−− 1
2
1
1
2
1
000
0000
00
α
α
α
α
[ ]XIY q 000 =
This is called the observable canonical form realization and has dimension rq.It is dual to (4.34).
验证式(4.33)具 如上 实 。这叫做能观 准型实 ，维 为 rq。它与式(4.34)对偶。
Answer:
)(
)(
1
)(
)(
1
,,
)(
,
)(
2,2,
)]([
][)(
1
11
2
2
1
1
1
1
1
1
1
1
21
21
1
r
r
qrq
r
q
r
r
i
i
i
q
r
i
iiq
i
iii
r
r
NNs
sd
BAsIC
then
I
sd
ZI
sd
s
ZI
sd
s
Z
then
s
Z
IZIsZ
ri
s
Z
ZriAZsZ
getwe
AsIZZZC
then
ZZZAsIC
define
++=−
===
−=−=
==⇒==
−=
=−
−−
−−
=
−
=
−
−
∑∑
α
α
this satisfies (4.33).
4.10 Consider the 21× proper rational matrix 考虑如下 则
[ ]
[ ]42322223124131221311
43
2
2
3
1
421
1
)(ˆ
bbbbbbbb
αααα
++++++×
++++
+=
ssssss
ssss
ddsG
Show that its observable canonical form realization can be reduced from Problem 4.9 as
试证习 4.9 统 能观 准形实 能 低维 如下表 ：
+
−
−
−
−
=
4241
3231
2221
1211
4
3
2
1
000
100
010
001
bb
bb
bb
bb
α
α
α
α
XX
[ ] [ ]UddXy 210001 +=
Answer: In this case,r=4,q=1,so
][],[],[],[,1 42414323132221212111 bbbbbbbb ===== NNNNI q
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its observable canonical form realization can be reduced from 4.9 to 4.10
4.11 Find a realization for the proper rational matrix 下 则 传递函 一 实 。
+
−−
−
+
−−++
=
+
+
−
+
−
++
−
+=
++
−
++
−
+=
11
00
26
34
23
22
23
1
11
00
2
2
1
3
)2)(1(
32
1
2
21
2
)2)(1(
32
1
2
)(ˆ
2
s
ss
ss
ss
s
s
s
s
s
s
ss
s
ssG
so the realization is
4.12 Find a
realization for
each column of
)(ˆ sG in Problem
4.11,and then
connect them.as shown in fig4.4(a).to obtain a realization of )(ˆ sG .What is the dimension of this
realization ?Compare this dimension with the one in Problem 4.11. 习 4.11 中 )(ˆ sG 一列
实 ，在如图 4.4(a)所 把它们连结得到 )(ˆ sG 一 实 。 较其与习 4.11 维 。
Answer:
222
222
2
111
111
1
1
0
22
32
0
1
01
23
1
0
2
3
2
2
)2)(1(
1
1
0
22
32
)2)(1(
1
)(ˆ
1
0
3
2
1
0
3
2
1
1
2
2
1
1
)(ˆ
UXY
UXX
s
sss
s
ss
sG
UXY
UXX
sss
sG
C
C
+
−−
−
=
+
−−
=
+
−
−
+
−++
=
+
−−
−
++
=
+
−
=
+−=
+
−+
=
−+
=
These two realizations can be combined as
UXY
UXX
+
−−−−
−
=
+
−−
−−
=
11
00
2623
3422
00
00
10
01
0010
0001
2030
0203
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UXY
UXX
+
−−−
−
=
+
−−
−
=
11
00
223
322
00
10
01
010
230
001
the dimension of the realization of 4.12 is 3,and of 4.11 is 4.
4.13 Find a realization for each row of )(ˆ sG in Problem 4.11 and then connect them,as shown in
Fig.4.4(b),to obtain a realization of )(ˆ sG .What is the dimension of this realization of this
realization?Compare this dimension with the ones in Problems 4.11 and 4.12. 习 4.11 中
)(ˆ sG 一行 实 ，在如图 4.4(b)所 把它们连结得到 )(ˆ sG 一 实 。 较其与习
4.11、4.12 维 。
Answer:
[ ] [ ] [ ][ ]
[ ] [ ]
[ ] [ ] [ ] [ ][ ] [ ]
[ ] [ ] 222
222
2
111
111
21
1101
26
23
02
13
112623
)2)(1(
1
11)1(2)2(3
)2)(1(
1
)(ˆ
0001
34
22
02
13
3422
23
1
3242
)2)(1(
1
)(ˆ
UXY
UXX
s
ss
ss
ss
sG
UXY
UXX
s
ss
ss
ss
sG
C
C
+=
−−
−−
+
−
−
=
+−−+−−
++
=++−+−
++
=
+=
−
+
−
−
=
−+
++
=−+
++
=
These two realizations can be combined as
UXY
UXX
+
=
−−
−−
−
+
−
−
−
−
=
11
00
0100
0001
26
23
34
22
0200
1300
0002
0013
the dimension of this realization is 4,equal to that of Problem 4.11.so the smallest dimension is
of Problem 4.12.
4.14 Find a realization for
+
+
+
+−
=
343
2322
343
)612(
)(ˆ
s
s
s
s
sG
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+
+
+
+−
=
343
2322
343
)612(
)(ˆ
s
s
s
s
sG 一 实
Answer:
[ ] [ ]
[ ]
[ ]UXY
UXX
ss
s
s
s
sG
3/224
9/6793/130
3
34
9/6793/130
3/34
1
3/224
343
2322
343
)612(
)(ˆ
−+=
−+−=
−
+
+−=
+
+
+
+−
=
4.16 Find fundamental matrices and state transition matrices for 下列 态 基
和 态转 ：
X
e
XandX
t
X
t
−
−
=
=
10
1
0
10 2
Answer:
for the first case:
25.0
22222
1
0
2121
)0()()(ln
)0()()(
t
t
extxtdttxdtxx
xdttxtxxx
=⇒=⇒=
+=⇒= ∫
we have
=⇒
=
=⇒
= ∫
2
2
5.0
0
5.0
)(
1
0
)0(
0
1
)(
0
1
)0(
t
t
d
e
e
tXXandtXX
ττ
=
=Φ
=
−
−
−
∫∫∫
∫
)(5.0
5.05.0
1
5.0
0
5.0
5.0
0
5.0
0
5.0
0
5.0
2
0
2
0
22
0
2
0
0 2
2
2
2
2
0
1
0
1
0
1
),(
0
1
)(
t.thusindependenlinearly are states initial twoThe
tt
t
t
ddt
t
t
d
t
t
d
t
t
d
e
ee
e
e
e
e
tt
and
e
e
tX
τττττττ
ττ
for the second case:
t
ttt
t dtt
dt
tt
extxxx
exeexcdteexetx
xextxexx
t
−
−−−
=⇒−=
+−=+∫∫=
⇒+−=+−=
∫
)0()(
)0())(0(5.0])0([)(
)0()(
2222
12
0
21
212
2
11
0
we have
−
=⇒
=
=⇒
=
−
−−
t
ttt
e
ee
tXXand
e
tXX
)(5.0
)(
1
0
)0(
0
)(
0
1
)0(
the two initial states are linearly independent;thus
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−+−
=
−
−
=Φ
−
=
−
−−−−
−
−
−−
−
−−
−
−−
tt
tttttttt
t
ttt
t
ttt
t
ttt
e
eeeeeee
e
eee
e
eee
tt
and
e
eee
tX
0
0000
0
000
0
)(5.0)(5.0
)(5.0
0
)(5.0
0
),(
)(5.0
0
)(
3
1
0
4.17 Show ).(),(/),( 00 tAttttt Φ−=∂Φ∂
试证 ).(),(/),( 00 tAttttt Φ−=∂Φ∂
Proof: first we know
)(),(/),(
0
),(
),()(
),(
),(),(),()(
),(
),(),(
),()),(),((),(
),()(
),(
00
0
0
10
0
1
00
0
00
000
0
0
tAttttt
t
tt
tttA
t
tt
tttttttA
t
tt
tttt
t
tt
t
tttt
t
tt
then
tttA
t
tt
Φ−=∂Φ∂⇒
=
∂
Φ∂
Φ+=
∂
Φ∂
Φ+ΦΦ=
∂
Φ∂
Φ+Φ
∂
Φ∂
=
∂
ΦΦ∂
=
∂
Φ∂
Φ=
∂
Φ∂
−−
4.18 Given
=
)()(
)()(
)(
2221
1211
tata
tata
tA ,show
+=Φ ∫ τττ daatt
t
t
))()((exp),(det 22110
0
已 A(t),试证
+=Φ ∫ τττ daatt
t
t
))()((exp),(det 22110
0
Proof:
τττ
τττ
φφφφ
φφφφφφφφφφφφ
φφφφφφφφφφφφ
φφ
φφ
φφ
φφ
daa
daa
t
t
t
t
ettthen
c
getwe
Ittwithcett
so
aaaa
aaaaaaaa
tt
so
aa
aa
∫=Φ
=
=Φ∫=Φ
Φ+=−+=
+−+++−+=
−+−=−
∂
∂
=Φ
∂
∂
=
+
+
0
2211
0
2211
))()((
0
00
))()((
0
2211122122112211
2122112112222212211121221212112221121111
122122111221221112212211
2221
1211
2221
1211
2221
1211
),(det
1
),(),(det
det)())((
)()()()(
)()(det
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4.20 Find the state transition matrix of 态转
x
tt
x
−
−
=
cos0
0sin.
Answer:
t
t
extxtxx
extxxtx
sin
2222
1cos
1111
)0()(cos
)0()(sin
−
−
=⇒−=
=⇒⋅−=
then we have
=⇒
=
=⇒
= −
+−
t
t
e
tXXand
e
tXX
sin
cos1 0
)(
1
0
)0(
0
)(
0
1
)0(
the two initial states are linearly independent;thus
=
=Φ
=
+−
−−
−
+−
−
+−
−
+−
0
0
0
0
sinsin
coscos
1
sin
cos1
sin
cos1
0
sin
cos1
0
0
0
0
0
0
),(
0
0
)(
tt
tt
t
t
t
t
t
t
e
e
e
e
e
e
tt
and
e
e
tX
4.21 Verify that
BtAtCeetX =)( is the solution of 试验证 BtAtCeetX =)( 为下 解：
CxXBAXX =+= )0(
Verify:first we know AeAee
t
AtAtAt ==
∂
∂
,then
DEQ
CCeeX
XBAXBeCeCeAeCee
t
X BtAtBtAtBtAt
..
)0(
)()()(
00 ==
+=+=
∂
∂
=
4.23 Find an equivalent time-invariant state equation of the equation in Problem 4.20.
习 4.20 中 价 不变 态 。
Answer: let )()()(
0
0
)()(
sin
cos1
1 txtPtxand
e
e
tXtP
t
t
=
==
−
−
Then 0)(0)( == txtA
4.24 Transform a time-invariant ),,( CBA into ))(),(,0( tCtB is by a time-varying equation
transformation. 把(A,B,C)转换成 ))(),(,0( tCtB 变 态转 。
Answer: since (A,B,C)is time-invariant,then
AtetX =)(
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)(
0
1
0
1
1
0)()(),(
)()(
)()(
)()(
ttA
At
At
etXtXtt
CetCXtC
BeBtXtB
tXtP
−−
−−
−
==Φ
==
==
=
4.25 Find a time-varying realization and a time-invariant realization of the inpulse response
tettg λ2)( = . 冲击响应 变实 及 不变实 。
Answer:
[ ]
−=
+−=−=−=
−
−
−
−−
λτ
λτ
λτ
λλλ
λτλτλ
τ
τ
τττττ
e
e
e
eteet
ettettgtg
ttt
tt
2
2
22)(2
2
)2()()(),(
the time-varying realization is
[ ] )(2)(
)()(
000
000
000
)(
2
2
txeteetty
tu
et
te
e
txtx
ttt
t
t
t
λλλ
λ
λ
λ
−=
+
=
−
−
−
[ ]xy
tuxx
nrealizatioinvianttimethegetwegu
sss
etLsg t
200
)(
0
0
1
010
001
33
:),41.4(sin
33
2
][)(ˆ
32
323
2
=
+
−
=
−
−+−
==
λλλ
λλλ
λ
4.26 Find a realization of ττ τ cos)(sin),( )( −−= tettg . Is it possible to find a time-invariant
state equation realization? ττ τ cos)(sin),( )( −−= tettg 一 实 ， 能否找到一 不变
态 实 。
Answer: clearly we can get the time-varying realization of
)()()cos)((sincos)(sin),( )( ττττ ττ NtMeteettg tt === −−− using Theorem 4.5
we get the realization.
)(sin)(
)(cos)()()(
txtety
ttuetutNtx
t
t
−=
==
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but we can’t get g(t) from it,because )(),( ττ −≠ tgtg ,so it’s impossible to find a
time-invariant state eqution realization.
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5.1
Is the network shown in Fing5.2 BIBO stable ? If not , find a bounded input that will excite an unbound output
图 5.2 网络 BIBO 否 定?如 不 .请举出 一般 励 输出 输入.
From Fig5.2 .we can obtain that
xy
yux
=
−=
Assuming zero initial and applying Laplace transform then we have
1)(ˆ
)(ˆ
)(ˆ
2 +
==
s
s
su
sy
sy
t
s
s
LsgLtg cos]
1
[)]([)(
2
11 =
+
== −−
because
∫ ∫ ∫ ∑∫
∞ ∞ ∞
=
π
+
π
+
+−+==
0 0
2
0
0
2
32
2
12
1 cos)1(cos|cos||)(|
x
k
k
k
k tdttdtdttdttg
=1+∑
∞
=
+ π+π−π+π−
0
2
1
2
31 )sin()2[sin()1(
k
k k
=1+∑
∞
=
+ π−π−−
0
2
1
2
31 ]sin[sin)1()1(
k
kk
=1+∑
∞
=
++ −⋅−−
0
1212 2)1()1(
k
kk ）（
=1+2∑
∞
=0
1
k
=∞
Which is not bounded .thus the network shown in Fin5.2 is not BIBO stable ,If u(t)=sint ,then we have
y(t)= ∫ ∫ τ−τ=τττ−
t t
tdt
0 0
)sin(cossin)cos( τd = ∫ =τ
t
tttd
0
sin
2
1
sin
2
1
we can see u(t)is bounded ,and the output excited by this input is not bounded
5.2
consider a system with an irrational function )(ˆ sy .show that a necessary condition for the system to be BIBO stable is
that |g(s)| is finite for all Res≥ 0
一个 统 传递函 S 式。证 该 统 BIBO 定 一个必要 件 对所 Res≥ 0. |g(s)| 。
Proof let s= ,ζ+λ j then
∫
∞ ζ−λ− ==ζ+λ=
0
)()(ˆ)(ˆ dteetgjgsg tjt ∫ ∫
∞ ∞ λ−λ− ζ−
0 0
sin)(cos)( tdtetgjtdtetg tt
If the system is BIBO stable ,we have
+∞<≤∫
∞
0
|)(| Mdttg . for some constant M
which implies shat ∫
∞
0
)( dttg is convergent .
For all Res= ,0≥λ We have
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|g(t) |cos te t ζλ− |)(| tg≤
|g(t0 |)(||sin tgte t ≤ζλ−
so ∫
∞ λ−
0
cos)( tdtetg t and ∫
∞ λ− ζ
0
sin)( tdtetg t are also convergent , that is .
∫
∞ λ− +∞<=ζ
0
cos)( Ntetg t
∫
∞ λ− +∞<=ζ
0
sin)( Ltdtetg t
where N and L are finite constant. Then
| 2
1
222
1
22 )())((|)(ˆ LNLNsg +=−+= is finite ,for all Res 0≥ .
Another Proof, If the system is BIBO stable ,we have .|)(|
0
+∞<≤∫
∞
Mdttg for some constant M .And
.
For all Res 0≥ . We obtain | =|)(ˆ sg ∫ ∫
∞ ∞− +∞<≤≤
0 0
)(Re |0(|)(| Mdttgdtetg ts .
That is . | |)(ˆ sg is finite for all Res 0≥
5.3
Is a system with impulse g(t)=
1
1
+t
BIBO stable ? how about g(t)=t
te− for t ?0≥
脉冲响应为个 g(t)=
1
1
+t
统 否 BIBO 定? 0,)( ≥= − ttetg t 统又如何?
Because ∞=+=
+
=
+∫ ∫
∞ ∞
)1ln(
1
1
|
1
1
|
0 0
ttd
t
dt
t
∫ ∫
∞ ∞ ∞−−−− +∞<=−−==
0 0
0 1|)(||
tttt etedttedtte
the system with impulse response g(t)=
1
1
+t
is not BIBO stable ,while the system with impulse response g(t)=
tte−
for 0≥t is BIBO stable.
5．4 Is a system with transfer function
)1(
)(ˆ
2
+
=
−
s
e
sg
s
BIBO stable
一个 统 传递函 为
)1(
)(ˆ
2
+
=
−
s
e
sg
s
， 该 统 否 BIBO 定。
Laplace transform has the property .
If ).()( atftg −= then )(ˆ)(ˆ sfesg sα−=
We know 0,]
1
1
[1 ≥=
+
−− te
s
L t . Thus 2,]
1
1
[ )2(1 ≥=
+
−−α−− te
s
eL ts
So the impulse response of the system is
)2()( −−= tetg . for 2≥t
∫∫
∞ −−∞ +∞<===
2
)2(
0
1|)(| dtedttg t
the system is BIBO stable
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5.5 Show that negative-feedback shown in Fig . 2.5(b) is BIBO stable if and only if and the gain a has a
magnitude less than 1.For a=1 ,find a bounded input )(tr that will excite an unbounded output.
证 图 2.5(b) 中 负反馈 统 BIBO 定当且增到 a 小于 1。对于 a=1 情况，找出一 输入 r(t), 她所产
输出 。
Poof ,If r(t)= )(td .then the output is the impulse response of the system and equal
+−d−−d−−d= )4()3()1()( 43 taratatg =∑
∞
=
+ −d−
1
1 )()1(
t
ii ita
the impulse is definded as the limit of the pulse in Fig.3.2 and can be considered to be positive .thus we have
∑
∞
=
−d=
1
)(|||)(|
t
i itatg
and ∫ ∑ ∫ ∑
∞ ∞
=
∞ ∞
=
∞
∞+<
−
==−d=
0
1
0
1 |)|1(
||||)(|||)(|
t t a
a
ii adtitadttg
1||
0||
<
≥
aif
aif
which implies that the negative-feedback system shown in Fig.2.5(b) is BIBO stable if and only if the gain a has
magitude less than 1.
For 1=a .if we choose ).sin()( π= ttr clearly it is bounded the output excited by )sin()( π= ttr is
∑
∫ ∑ ∑ ∑ ∑
∞
=
∞
+
∞
=
∞
=
∞
=
++++
π−=
π−−=π−π−=π−π−=π−π−=τττ−=
1
0
1 1 1 1
1111
1)sin(
)sin()1()1()sin()1()sin()1()sin()1()()()(
i
t
i i i i
iiiii
t
titititdrtgty
And y(t) is not bounded
5.6 consider a system with transfer function
)1(
)2(
)(ˆ
+
−
=
s
s
sg 。 what are the steady-state responses by
u( t )=3.for 0≥t .and by u ( t )=sin2t. for ?0≥t
一个 统 传递函
)1(
)2(
)(ˆ
+
−
=
s
s
sg ，分别 0.3)( ≥= ttu 和 0.2sin)( ≥= tttu
所产 态响应 什么？
The impulse response of the system is 0.3)(]
1
3
[`]
1
2
[)( 11 ≥−d=
+
−=
+
−
= −−− tet
s
L
s
s
Ltg t 。
And we have
∫ ∫∫ ∫ ∫
∞ ∞ −∞ ∞ ∞ −− +∞<=+=+d=+d≤−d=
0 00 0 0
4313)()3||)((||3)(||)(| dtedttdtetdtetdttg ttt
so the system is BIBO stable
using Theorem 5.2 ,we can readily obtain the steady-state responses
if u ( t )=3for 0≥t .,then as )(, tyt ∞→ approaches 6323)0(ˆ −=⋅−=⋅g
if u ( t )=sin2t for ,0≥t then ,as )(. tyt ∞→ approaches
)25.1sin(26.1)3arctansin(
5
102
)(ˆsin(|)(ˆ| +τ=+τ=τ∠+ττ ttjgtjg ）
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5.7 consider =x ux
−
+
−
0
2
10
101
y=[ -2 3] ux 2− is it BIBO stable ?
上述 态 述 统 否 BIBO 定？
The transfer function of the system is
[ ] [ ] 2
0
2
1
1
0
)1)(1(
10
1
1
322
0
2
10
101
32)(ˆ
1
−
−
−
−++−=−
−
−≤
−+
−=
−
s
sssssg
=
1
22
2
1
4
+
+−
=−
+ s
s
s
The pole of is –1.which lies inside the left-half s-plane, so the system is BIBO stable
5.8 consider a discrete-time system with impulse response
kkkg )8.0(][ = for ,0≥k is the system BIBO stable ?
一个 统 脉冲响应序列
kkkg )8.0(][ = ， ,0≥k 。 该 统 否 BIBO 定？
Because
∑ ∑ ∑
∞
=
∞
=
∞
=
⋅−==
0 0 0
)8.0()8.01(
2.0
1
)8.0(|][|
k k k
kk kkkg =
∑ ∑∑
∞
=
∞
=
∞
=
+ +∞<=
−
⋅==⋅−⋅
0 00
1 20
8.01
8.0
2.0
1
8.0
2.0
1
])8.0()8.0([
2.0
1
k k
k
k
kk kk
G[k]is absolutely sumable in ∞,0[ ]. The discrete-time system is BIBO stable .
5.9 Is the state equation in problem 5.7 marginally stable? Asymptotically stable ? 5.7 中 态
述 统 否 定？ 否 进 定？
The characteristic polynomial
)1)(1(
10
101
det)det()( −λ+λ=
−λ
−+λ
=−λ=λ∆ AI
the eigenralue Iֹhas positive real part ,thus the equation is not marginally stable neither is Asymptotically stable.
5.10 Is the homogeneous stable equation xx
−
=
000
000
101
.marginally stable ? Asymptotically stable?
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齐 态 xx
−
=
000
000
101
， 否 定？
The characteristic polynomial is )1(
00
00
101
det)det()( 2 +λλ=
λ
λ
−+λ
=−λ=λ∆ AI
And the minimal polynomial is )1()( +λλ=λψ .
The matrix has eigenvalues 0 , 0 . and –1 . the eigenvalue 0 is a simple root of the minimal .thus the equation is marginally
stable
The system is not asymptotically stable because the matrix has zero eigenvalues
5 。11 Is the homogeneous stable equation xx
−
=
000
000
101
marginally stable ? asymptotically stable ?
齐 态 xx
−
=
000
000
101
否 定？ 否 进 定？
The characteristic polynomial ）（ 1
00
10
101
det)( 2 +λλ=
λ
−λ
−+λ
=λ∆ .
And the minimal polynomial is )1()( 2 +λλ=λψ ..the matrix has eigenvalues 0. 0. and –1 . the eigenvalue 0 is a
repeated root of the minimal polymial ).(λΨ so the equation is not marginally stable ,neither is asymptotically
stable .
5.12 Is the discrete-time homogeneous state equation
][
100
010
109.0
]1[ kxkx
−
−
=+
marginally stable ?Asymptotically stable ?
齐 态 ][
100
010
109.0
]1[ kxkx
−
−
=+ 否 定？ 否 进 定？
The characteristic of the system matrix is ),9.01()( 2 −λ−λ=λ∆ （） .and the minimal polynomial is
),9.01()( −λ−λ=λψ （） the equation is not asymptotically stable because the matrix has eigenvalues 1.
whose magnitudes equal to 1 .
the equation is marginally stable because the matrix has all its eigenvalues with magnitudes less than or to 1 and
the eigenvalue 1 is simple root of the minimal polynomial . )(λψ .
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5.13 Is the discrete-time homogeneous state equation
][
100
010
109.0
]1[ kxkx
−
−
=+ marginally stable ?
asymptotically stable ?
齐 态 ][
100
010
109.0
]1[ kxkx
−
−
=+ 否 定？ 否 进 定？
Its characteristic polynomial is ),9.01()( 2 −λ−λ=λ∆ （） and its minimal polynomial is
),9.01()( 2 −λ−λ=λψ （） the matrix has eigenvalues 1 .1 .and 0.9 . the eigenvalue 1 with magnitude equal to
1 is a repeated root of the minimal polynimial )(λψ . .thus the equation is not marginally stable and is not
asymptotically stable .
5.14 Use theorem 5.5 to show that all eigenvalues of
−−
=
15.0
10
A have negative real parts .
定 5.5 证 A 征值都具 负实部.
Poof :For any given definite symmetric matrix N where
=
cb
ba
N with a>0 .and 02 >− bac
The lyapunov equation NMAMA −=+′ can be written as
−−
−−
=
−−
+
−
−
cb
ba
mm
mm
mm
mm
15.0
10
11
5.00
2221
1211
2221
1211
that we have
−−
−−
=
−+−−
−−+−
cb
ba
mmmmmm
mmmmm
222112222111
2212112112
25.0
5.0)(5.0
thus
cam
amm
bcam
5.0
25.05.1
22
2112
11
+=
==
−+=
=
2221
1211
mm
mm
M is the unique solution of the
lyapunor equation and M is symmetric and
0
0
2 >−
>
bac
a
⇒
0
0
>
>
c
a
025.05.1
0,0
)25.05.1(
0]32)2[(
16
1
)436(
16
1
]16)6[(
16
1
)25.05.1(
11
22
22222
>−+=⇒
>>
>≥+
≥+−=−+=−+=−+∴
bcam
ca
bacca
acaaccaaccaacca
2
21122211
2221
1211
)5.)(25.05.1(det acabcammmm
mm
mm
−−+−+=−=
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0)](8)2()22[(
8
1
)4884(
8
1
222
22
>−+−+−=
−−++=
bacbcba
bcabcaca
we see is positive definite
because for any given positive definite symmetric matrix N, The lyapunov equation NMAMA −=+′ has a
unique symmetric solution M and M is positive definite , all eigenvalues of A have negative real parts ,
5.15 Use theorem 5.D5show that all eigencalues of the A in Problem 5.14 magnitudes less than 1.
Poof : For any given positive definite symmetric matrix N , we try to find the solution of discrete Lyapunov
equation M
NMAAM =′− Let .
=
cb
ba
N where a>0 and 02 >− bac
and M is assumed as
=
2221
1211
mm
mm
M ,then
=
++−−+
−+−
=
−−
−
−
−
=′−
cb
ba
mmmmmm
mmmmm
mm
mm
mm
mm
MAAM
211211221221
2221122211
2221
1211
2221
1211
)50
)(5.025.0
15.0
10
11
5.00
（。
this bcambcammbcam
5
16
5
12
5
12
,
5
2
5
4
5
4
,
5
4
5
3
5
8
22211211 −+=−+==−+=
=
2221
1211
mm
mm
M is the unique symmetric solution of the discrete Lyapunov equation
,NMAAM =′− And we have
>
>
⇒
−
>
++=−+
0
00
3296416)38(
2
222
ca
a
bac
a
accaacca
bacca 1616)38( 2 >>+⇒ 0438 >−+⇒ bca 0
5
4
5
3
5
8
11 >−+=⇒ bcam
0)](5)2()22[(
5
4
)48534(
5
4
])244()161212)(4(38[(
25
1
det
222
222
2
21122211
2221
1211
>−+−+−=
−−+++=
−+−−+−+=−=
bacbcba
bcabacbca
bcabcabammmm
mm
mm
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we see M is positive definite .use theorem 5.D5,we can conclude that all eigenvalues of A have magnitudes less
than 1.
5.16 For any distinct negative real iλ and any nonzero real ia ,show that the matrix
λ
−
λ+λ
−
λ+λ
−
λ+λ
−
λ
−
λ+λ
−
λ+λ
−
λ+λ
−
λ
−
=
3
2
3
23
23
13
13
32
32
2
2
2
12
12
31
31
21
21
1
2
1
2
2
2
aaaaa
aaaaa
aaaaa
M is positive definite .
上定义 M 宗 （ iλ 为 异 负实 ， ia 为 实 ）。
Poof : .3,2,1, =λ ii are distinct negatire real numbers and ia are nonzero real numbers , so we have
0
12
),1(
2
1 >
λ
−
a
(2) . ）
2
2121
2
2
2
12
21
2
2
2
1
21
2
2
2
1
2
2
2
12
12
21
21
1
2
1
)(
1
4
1
(
)(4
2
2
det
λ+λ
−
λλ
=
λ+λ
−
λλ
=
λ
−
λ+λ
−
λ+λ
−
λ
−
aa
aaaa
aaa
aaa
2
2121
21
2
21
2
212121
2
2
2
121
2
21
)(
1
4
1
4
424
λ+λ
>
λλ
λλ>λ+λ∴
λ−λ=λλ−λλ+λ+λ=λλ−λ+λ
and）（
〉）（）（
λ
−
λ+λ
−
λ+λ
−
λ
−
2
2
2
12
12
21
21
1
2
1
2
2
det
aaa
aaa
=
2
2121
2
2
2
1
)(
1
4
1
(
λ+λ
−
λλ
aa ) >0
(3)
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λ+λ
λ
−
λλ+λλ+λ
λ
−
λ+λ
λ+λλ+λ
λ
−
λ+λλ+λ
λ
−
λ
λ+λλ+λλ
−=
λ
−
λ+λ
−
λ+λ
−
λ+λ
−
λ
−
λ+λ
−
λ+λ
−
λ+λ
−
λ
−
−=
2
21
1
33121
1
32
3121
1
32
2
12
3
2
31211
2
3
2
2
2
1
32313
32212
31211
2
3
2
2
2
1
2
2
1
))(
21
0
))(
212
2
1
0
11
2
1
det
2
111
1
2
11
11
2
1
det
）（（
（）（
）
）
aaa
aaaM
0det
.0
))(
4
,0
)(
,0.0
4
1
,0
2
1
))(
4
)(
1
)(4
1
det
2
1
))(
212
2
12
2
1
det
2
1
3213121
1
312
1
213
1
213
1
32
2
3
2
2
2
1
1
3213121
1
2
32312
1
213
1
32
2
3
2
2
2
1
1
2
3121
1
32
2
13
1
3
2
12
1
2
2
3
2
2
2
1
1
<∴
>
λ+λλ+λλ+λ
λ
>
λ+λλ
λ
−>
λ+λλ
λ
−>
λ+λλ
λ
−
λλ
>
λ
−
λ+λλ+λλ+λ
λ
+
λ+λ
−
λ+λλ
λ
−
λ+λλ
λ
−
λλλ
−=
λ+λλ+λ
λ
−
λ+λ
−
λ+λ
λ
−
λλ+λ
λ
−
λλ
−=
=
=
M
aaa
aaa
aaa
）（）（
）（）（
）（）（）（
）
）（
（）
）（
）（
）（
（
From (1), (2) and (3) , We can conclude that M is positive definite
5.17 A real matrix M (not necessarily symmetric ) is defined to be positive definite if 0>′ XMX
for any non zero X .Is it true that the matrix M is positive definite if all eigenvalues of Mare real
and positive or you check its positive definiteness ?
一个实 M 被定义为 定 ，如 对任意 X 都 0>′ XMX 。如 M 所 征值都
实 ， 不 就 M 定/ 或者如 M 所 主子式都 ，M 不 定呢？如 不
，要如何 定它 定性呢？
If all eigenvalues of M are real and positive or if all its leading pricipal minors are positive ,the matrix
M may not be positive definite .the exampal following can show it .
Let
−
=
10
81
M , Its eigenvalues 1,1 are real and positive and all its leading pricipal minors are
positive , but for non zero [ ] ′= 21X we can see 011<−=′MXX . So M is not positive
definite .
Because MXX ′ is a real number . we have XMXXMXXMX ′=′′=′′ )( , and
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XMMXXMXXMXXMX )]
2
1
[
2
1
2
1 ′+′=′′+′=′ （
This ifonlyandifXMX ,0>′ XMMX )]
2
1
[ ′+′ （ >0, that is , M is positive and the matrix
)
2
1
MM ′+（ is symmetric , so we can use theorem 3.7 to cheek its positive definiteness .
5.18 show that all eigenvalues of A have real parts less than 0<µ− if and only if ,for any given
positive definite symmetric matrix N , the equation NMMAMA −=µ++′ 2 has a unique
symmetric solution M and M is positive definite .
证 A 所 征值 实部小于 0<µ− 当且对任意给定 定对 N。
NMMAMA −=µ++′ 2 唯一解 M，且 M 定 。
Proof : the equation can be written as NIAMIA −=µ++′µ+ )()(
Let IAB µ+= , then the equation becomes NMBMB −=+′ . So all eigenvalues of B have
megative parts if and only if ,for any given positive definite symmetric matrix N, the equation
NMBMB −=+′ has a unique symmetric solution m and is positive definitive .
And we know IAB µ+= , so ( )AIIAIBI −µ−λ=µ−−λ=−λ )(det)det()det( , that is , all
eigenvalues of A are the eigenvalues of B subtracted µ− . So all eigenvalues of A are real parts less
than 0<µ− .if and only all eigenvalues of B have negative real parts , eguivalontly if and only if ,
for any given positive symmetric matrix N the equation NMMAMA −=µ++′ 2 has a unique
symmetric solution M and M is positive definite .
5.19 show that all eigenvalues of A have magnitudes less than r if and only if , for any given
positive definite symmetric matrix N ,the equation NMAAM 22 r=′−r . Has a unique symmetric
solution M and M is positive definite .
证 A 所 征值 都小于 r 当且仅当对任意给定 对 定 N ，
NMAAM 22 r=′−r 唯一解 M ，且 M 对 定 。
Proof : the equation can be written as NAMcM =r′− − )()( 1 let AB 1−r= , then
).det()det()det( 11 AIAIBI −rλr=r−λ=−λ −− that is ,all eigenvalues of A are the eigenvalues
B multiplied by r ,so all eigenvalue of B have msgritudes less than 1 . equivalently if and only if ,
for any given positive definite symmetric matrix N the equation NMAAM 22 r=′−r has a unique
symmetric solution M and M is positive definite .
5．20 Is a system with impulse response ||||2),( τ−−=τ tetg , for τ≥t , BIBO stable ? how about
?cos)sin(),( )( tetg t τ−−=τ
脉冲响应为
||||2),( τ−−=τ tetg ， τ≥t ， 统 否 BIBO 定？ ?cos)sin(),( )( tetg t τ−−=τ
统 否 BIBO 定？
τ=τ=τ ∫∫∫ −−−−− dedede
t
t
t
t
t
tt
t
t
tt
000
||||||2||||2 || 、
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<<−−
><−−
≥−
=
−
−−
−−−
00)2(
00)2(
0)(
0
2
0
2
0
2
0
0
0
tandtifeee
tandtifeee
tifeee
ttt
ttt
ttt
∞+<≤ 2
)1(|sin||cos|sin||cos)sin(| 0)()(
tttt ettette
−τ−−τ−− −⋅≤⋅⋅=
∫ ∫ −−τ−τ−− −⋅=τ⋅=τ⋅
t
t
tt
t
t
tt etdeetdet
0
0
0
)1(|sin||sin|)(|sin|
)()(
]1,0(, 00 ∈∴≤
−tt
ett
∫ +∞<≤τ⋅ τ−−
t
t
t sodet
0
0)(|sin| )(
∫ ∫ +∞<≤⋅⋅≤τ⋅ −−−τ−−
t
t
tt
t
t
tt etetet
0
0
0
1|sin||sin||cos)(sin|
)()(
Both systems are BIBO stable .
5.21 Consider the time-varying equation
xteyutx
2
2 −=+τ= is the equation stable ?
marginally stable ? asymptotically stable ?
变 xteyutx
2
2 −=+τ= 否 BIBO 定 定？ 进 定？
22 )()0()()(20(
tt
etxextxttxtx =⇒=⇒= is a fundamental matrix
⇒ its stable transition matrix is 0
2
),( 0
tt
ett
−=φ
⇒ Its impulse response is
2222
),( τ−τ−− =⋅=τ eeetg tt
+∞<π=
π
⋅=τ=τ=⋅τ≤τ=ττ ∫∫ ∫∫
∞+ τ−τ−∞+
∞−
τ−τ−
2
22),(
0
22
0
22
0
dededededtg
t
t
t
t
so the equation is BIBO stable .
00 ,|||),(|
0
2
0
2
tteett
tttt ≥==φ −− is not bounded ,that is . there does not exist a finite
constant M such that .),( 0 +∞<≤φ Mtt so the equation is not marginally stable and is not
asymptotically stable .
5.22 show that the equation in problem 5.21 can be transformed by using
,)( xtpx = with
2
)( tetp −= , into xyuexx t =+= −
2
.0 is the equation BIBO
stable ?marginally stable ? asymptotically ? is the transformation a lyapunov transformation ?
证 5 。 21 可 通 过 ,)( xtpx = with
2
)( tetp −= 变 换 成
xyuexx t =+= −
2
.0 该 否 BIBO 定？ 定？ 进 定？该变换 不
lyapunov 变换 ?
proof : we have found a fundamental matrix of the equation in problem 5.21 .is .)(
2tetx = so
from theorem 4.3 we know , let .)()
21 tetxtp −
−
==（ and ,)( xtpx = , then we have
0)( =tA
0)().1))().))()
2221
=======−−
−
tDtDeetBtCtCetBtxtB ttt （（（（（
and the equation can be transformed into xyuexx t =+= −
2
.0
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the impulse response is invariant under equivalence transformation , so is the BIBO stability .
that is marginally stable . 1)()0()(.0 =⇒=⇒→= txxtxxx is a fundamental matrix
⇒ the state transition matrix is +∞<≤=φ=φ 11|),(|1),( 00 tttt so the equation is
marginally stable . 0|),(|, 0 →φ∞→ ttt (不趋于 )so the equation is not asymptotically
stable . from theorem 5.7 we know the transformation is mot a lyapunov transformation
5.23 is the homogeneous equation 0
0
01
03
≥
−
−
= − tforxe
x
t
marginally stable ? asymptotically stable?
齐 0,
0
01
03
≥
−
−
= − txe
x
t
否 定？ 进 定？
=⇒
−+=
=
⇒
−=
−=
⇒
−
−
=
−
−
−
−
−−
1
4
1
0
)0(
4
1
)0()0(
4
1
)(
)0()(
0
01
4
12
4
12
11
1
3
2
11
3
t
t
t
t
tt
e
e
x
xxextx
extx
xex
xx
x
e
x
is a fundamental matrix of the equation .
==Φ⇒ +−
+−
−
1
4
1
0
)()()(
0
0
40
1
tt
tt
e
e
tXtXt is the state
transition of the equation.
−+=φ −+−+−∞ 000
34
0
4
1
4
1
1,max||),(||
ttttt
eeett for all 00 ≥t
and 0tt ≥
4
1
4
1
0;10 00
4 ≤≤≤≤ +−+− tttt ee soee ttt ,0
4
1
4
1
;10 00
3 ≤≤−≤≤ −+−
4
5
4
1
4
1
1
4
3
00 34 ≤−+≤ −+− ttt ee . +∞<≤φ ∞
4
5
||),(|| 0tt the equation is marginally
stable .if 0tt − is fixed to some constant , then 0||),(|| 0 →φ tt (不趋于 )。 ,∞→tas
so the equation is not asymptotically stable .
for all t and 0t with 0tt ≥ . we have
4
1
4
1
0;10 00
4 ≤≤≤≤ +−+− tttt ee . every entry of
),( 0ttφ is bounded , so the equation is marginally stable .
because the (2,2)th entry of ),( 0ttφ does not approach zero as ,∞→t the equation is not
asymptotically .
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6.1 is the stable equation
xy
xx
]121[
0
0
1
331
100
010
=
µ
+
−−−
=
controllable ? observable ?
述 态 否可控？ 否可观？
1
121
121
021
(
0
0
1
3
331
100
010
(])([ 2
=
−−−r=
r
=
−−−
r=r
））（
）BAABB
thus the state equation is controllable ,
but it is not observable .
6.2 is the state equation
xy
uxx
]101[
00
01
10
121
100
010
=
+
−−
=
controllable ? observable?
述 态 否可控？ 否可观？
,1232.2)( =−=−=r nB using corollary , we have
3)
0200
0001
0110
(])([ =
r=r ABB ,Thus the state equation is controllable
3)
420
130
101
()(
2
=
−
−r=
r
CA
CA
C
. Thus the state equation is observable
6.3 Is it true that the rank of [B AB ]1BAn− equals the rank of ][ 2 BABAAB n ? If
not ,under what condition well it be true ?
r ( [B A ][ 2 BABAAB n B ]1BAn− ]=r () 否成 ？若否， 在什么 件下该成
？
It is not true that the rank of [B AB ]1BAn− equals the rank of ][ 2 BABAAB n .
If A is nonsingular {( r ( ][ 2 BABAAB n )= r (A [B AB ]1BAn− ]= r ( [B
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AB ]1BAn− (see equation (3.62) ) then it will be true
6.4 show that the state equation u
B
x
AA
AA
x
+
=
0
1
2221
1211 is controllable if and only if the pair
)( 2122 AA IS controllable 证 述 态 可控当且仅当 )( 2122 AA 可控？
Proof :
6.5 find a state equation to describe the network shown in fig.6.1 and then check its controllability
and observability
图 6.1 中网络 一个 态 , 否可控?可观?
The state variables 1x and 2x are choosen as shown ,then we have
uxy
xx
uxx
2
0
2
22
11
+−=
=+
=+
thus
a state equation describing the network can be es pressed as
u
x
x
y
U
x
x
x
x
2]10[
0
1
10
01
2
1
2
1
2
1
+
−−
+
−
−
=
checking the controllability and observability :
1)
10
01
()(
1)
10
01
(]([
=
−
−
r=
r
=
−
−
r=r
CA
C
ABB
The equation is neither controllable nor obserbable
6.6 find the controllability index and obserbability index of the state equation in problems 6,1 and 6.2
6.1 和 6.2 中 态 可控性指 和可观性指 .
The state equation in problem 6.1 is
u
x
x
y
U
x
x
x
x
2]10[
0
1
10
01
2
1
2
1
2
1
+
−−
+
−
−
=
the equation is controllable ,so
we have 1)(,3.1)1,min( ===+−≤+−≤µ≤ brankpnwherepnpnnpn
333.1 =µ≤µ≤+−≤µ≤∴ iepnpn
[ ]( ) 1,1
2
=∴=
r
r=r V
AC
AC
C
AC
C
c
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The controllability index and cbservability index of the state eqyation in problem 6.1 are 3 and 1 ,
respectively .the state equation in problem 6.2 is
[ ]xy
uxx
101
00
01
10
120
100
010
−
+
−
=
The state equation in is both controllable and observable ,so we have
1.1 +−≤ν≤+−≤µ≤∴ pnpnandpnpn
1.2)(,3 ===== crankqBrankpnwhere
3,233.223 =ν=µ≤µ≤≤µ≤∴
The controllability index and cbservabolity index of the state equation in problem 6.2 are 2 and
3 ,respectively rank(c)=n .the state equation is controllable so
1)(.1 =µ⇒==+−≤µ≤ nBrankpwherepnpn
6.7 what is the controllability index of the state equation IUXAx += WHERE I is the unit matrix ?
态 IUXAx += 可控性指 什么？
Solution . ][][ 1212 −− == nn AAAIBABAABBC
C is an
2nn× matrix. So rank(C) n≤ . And it is clearly that the first n columns of C are
linearly independent. This rank(C) n≤ .and niforui ,1,01 ==
The controllability index of the state equation IUXAx += is 1),,,max( 21 == nuuuu
6.8 reduce the state equation xyuxx ]11[
1
1
13
41
=
+
−
−
= . To a controllable one . is the
reduced equation observable ?
将 述 态 可控 ， 维后 否可观？
Solution :because [ ]( ) 21)
33
31
()( <=
r=r=r ABBC
The state equation is not controllable . choosing
== −
01
11
1PQ and let xpx = then we
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have
−
=
−
−
−
== −
50
43
01
11
14
41
11
10
1PAPA
=
−
==
0
1
1
1
11
10
PBB . [ ] [ ]12
01
11
111 =
== −CPC
thus the state equation can be reduced to a controllable subequation
ccc xyuxx 23 =+= and this reduced equation is observable
6.9 reduce the state equation in problem 6.5 to a controllable and observable equation . 将 6.5 中
态 维为可控且可观 .
Solution : the state equation in problem 6.5 is
uxy
Uxx
2]10[
0
1
10
01
+−−
+
−
−
=
it is neither controllable nor observable
from the form of the state equation ,we can reaelily obtain
u
x
x
y
U
x
x
x
x
2]10[
0
1
10
01
2
1
2
1
2
1
+
−−
+
−
−
=
thus the reduced controllable state equation can be independentof cx . so the equation can be further
reduced to y=2u .
6.10 reduce the state equation
[ ]xy
uxx
10110
1
0
0
1
0
0000
1000
0000
0010
0001
2
2
1
1
1
=
+
λ
λ
λ
λ
λ
=
to a controllable and
observable equation 将 态 维为可控可观 。
Solution the state equation is in Jordan-form , and the state variables associated with 1λ are
independent of the state variables associated with 2λ thus we can decompose the state equation into
two independent state equations .and reduce the two state equation controllable and observable equation ,
respectively and then combine the two controllable and observable equations associated with 1λ and
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2λ . Into one equation ,that is we wanted .
Decompose the equation .
[ ]
[ ] 22
2
2
2
11
1
1
1
1
10
1
0
0
1
110
0
1
0
00
10
01
xyuxx
xyuxx
=
+
λ
λ
=
=
+
λ
λ
λ
=
Where 21
2
1
, yyy
x
x
x +=
=
Deal with the first sub equation .c
Choosing 111
1
100
01
010
: xpxletandPQ =
λ== − then we have
λ
λ
λ−
=
λ
λ
λ
λ
−
== −
1
1
1
2
1
1
1
1
1
11
00
021
10
100
01
010
00
10
01
100
001
01C
PPAA
=
λ−
==
0
0
1
0
1
0
100
000
0111
11 PBB [ ] [ ]11
100
01
010
110 1
1
11 λ=
== − CPCC
Thus the sub equation can be reduced to a controllable one
[ ]
21
1
)(
21
0
1
1
0
2
11
1
111
1
1
2
1
1
<=
λλ
λ
r=r
λ=
+
λ
λ−
=
O
xy
uxx
c
cc
the reduced controllable equation is not observable choosing
λ
=
10
1 1
P .and let ,
cc xpx 11 = then we have
[ ] [ ]01
10
1
1
0
1
0
1
10
1
1
0
10
1
21
0
10
1
1
11
1
1
1
11
1
2
11
1
=
λ−
λ=
=
λ
=
λ
λ
=
λ−
λ
λ−
λ
=
C
B
A
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thus we obtain that reduced controllable and observable equation of the sub equation associated
with
011
01111 :
c
cco
xy
uxx
=
+λ=λ
deal with the second sub equation :
2)
1
10
()(
2
2 =
λ
r=r C it is controllable
21)
0
10
()(
2
2 <=
λ
r=r O It is not observable choosing
=
01
10
p and let
22 xpx = then we have
[ ] [ ]01
01
10
10
0
1
1
0
01
10
1
0
01
10
0
1
01
10
2
2
2
2
2
2
2
=
=
=
=
λ
λ
=
λ
λ
=
C
B
A
thus the sub equation can be reduced to a controllable and observable
ico
icoico
xy
uxx
=
+λ=
2
2
combining the two controllable and observable state equations . we obtain the reduced controllable and
observable state equation of the original equation and it is
[ ] co
coco
xy
uxx
11
1
1
0
0
2
1
=
+
λ
λ
=
6.11 consider the n-dimensional state equation
uoxcy
uBxAx
+=
+=
the rank of its controllability matrix is
assumed to be nn <1 . Let 1Q be an 1nn× matrix whose columns are any 1n linearly independent
columns of the controllability matrix let 1p be an nn ×1 matrix such that 111 InQp = ,where 1In
is the unit matrix of order 1n .show that the following 1n -dimensional state equation
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uDxCQy
uBPxAQPx
+=
+=
11
11111
is controllable and has the same transfer matrix as the original state equation
n 维 态
uoxcy
uBxAx
+=
+=
。假设共可控性 为 nn <1 ，令 1Q 为可控性 中任意 1n 个
性 关 成 1nn× ，而 1p P 发 且 111 InQp = ，其中 1In 1n 单位 。证 下列 1n
维 态
uDxCQy
uBPxAQPx
+=
+=
11
11111
可控且与原 态 具 同 传函。
Let
== −
2
11 :
P
P
QP , where 1P is an nn ×1 matrix and 2P is an nnn ×− )( 1 one ,and we
have
1
1
1
1
22
11
2212
2111
2211 0
0
nn
n
n
nn
n
n IQP
IQP
isthatI
I
I
QPQP
QPQP
PQIPQPQQP
−− =
=
=
=
==+=
Then the equivalence trans formation xpx = will transform the original state equation into
[ ]
u
B
x
x
A
AA
u
Bp
Bp
x
x
AQPAQP
AQPAQP
uB
p
p
x
x
QQA
p
p
x
x
c
c
c
C
C
c
c
c
c
c
c
+
=
+
=
+
=
00
12
2
1
2212
2111
2
1
21
2
1
[ ]
[ ]
[ ] uD
x
x
CC
uD
x
x
CQCQ
uD
x
x
QQCy
c
c
cc
c
c
c
c
+
=
+
=
+
=
21
21
where 111111 ,.. nnisAandCPCBPBAQPA cCcc ×=== the reduced 1n -dimensional
state equation
uDxCQy
uBPxAQPx
+=
+=
11
11111
is controllable and has the sane transfer matrix as the original
state equation
6.12 In problem 6.11 the reduction procedure reduces to solving for 1P in .11 IQP = how do you
solve 1P ?
6.11 中 维过 为 解 .11 IQP = 中 1P , 该如何 1P ?
Solution : from the proof of problem 6.11 we can solve 1P in this way
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r (C)= r ][ 2 BABAAB n = nn <1
we fprm the nn < matrix ]1 1([ nn qqqQ = , where the first 1n columns are any 1n linearly
independent columns of C , and the remaining columns can aribitrarily be chosen as Q is nonsingular .
Let
== −
2
11 :
P
P
QP , where 1P is an nn ×1 matrix and 111 InQp =
6.13 Develop a similar statement as in problem 6.11 for an unobservable state equation .对一不可观
态 ，如 6.11 般低哦 维为可观 .
Solution : consider the n-dimensional state equation
uDxCy
uBxAx
+=
+=
the rank of its observability matrix
is assumed to be nn <1 .let 1P be an nn ×1 matrix whose rows are any 1n linearly independent
rows of the observability matrix . let 1Q be an 1nn× matrix such that 111 InQp = .where 1In is
the unit matrix of order 1n , then the following 1n -dimensional state equation
uDxCQy
uBPxAQPx
+=
+=
11
11111
is observable and has the same transfer matrix as the original state equation
6.14 is the Jordan-form state equation controllable and observable ?
下 Jordan-form 态 不 可控，可观？
uxx
−
+
=
001
101
101
123
111
112
012
1000000
0100000
0110000
0002000
0000200
0000020
0000012
xy
−
=
0110110
0002111
1113122
solution : 2)
001
101
(3)
123
111
112
( =
r=
r and
so the state equation is controllable ,however ,it is not observable because
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32)
110
211
312
( ≠=
r
6.15 is it possible to find a set of ijb and a set of ijc such that the state equation is
controllable ? observable ? 否可能找到一 ijb 和一 ijc 使如下 态 可控？可观？
u
ccccc
ccccc
ccccc
yu
bb
bb
bb
bb
bbxx
=
+
=
3534333231
2524232221
1514131211
5251
4241
3231
2221
1211
10000
01000
01100
00010
00011
solution : it is impossible to find a set of ijb such that the state equation is observable ,because
32(
5251
4241
2221
<≤
r ）
bb
bb
bb
it is possible to find a set of ijc such that the state equation is observable .because
3(
353331
252321
151311
≤
r ）
CCC
CCC
CCC
The equality may hold for some set of ijc such as I
CCC
CCC
CCC
=
353331
252321
151311
the state equation is
observable if and only if 3(
353331
252321
151311
=
r ）
CCC
CCC
CCC
6
6.16 consider the state equation
[ ]xCCCCCy
u
b
b
b
b
b
xx
222112111
22
21
12
11
1
12
22
11
11
1
000
000
000
000
0000
=
+
αb
bα
αb−
bα
λ
=
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it is the modal form discussed in (4.28) . it has one real eigenvalue and two pairs of complex
conjugate eigenvalues . it is assumed that they are distinct , show that the state equation is
controllable if and only if 2,100;0 211 =≠≠≠ iforborbb ii it is observable if
and only if 2,100;0 211 =≠≠≠ iforcorcc ii
中 态 （4.82）中讨论 规范型。它 一个实 征值和两对复 征值，并假定它
们都 互不 同 。证 该 态 可控当且仅当
2,100;0 211 =≠≠≠ iforborbb ii 可观当且仅当
2,100;0 211 =≠≠≠ iforcorcc ii
proof:；controllability and observability are invariant under any equivalence transformation . so we
introduce a nonsingular matrix is transformed into Jordan form.
−
−=
−
−
= −
ji
iip
j
j
j
j
p
000
11000
000
00110
00001
,
5.05.0000
5.05.0000
005.05.00
005.05.00
00001
1
+
−
+
−
==
b−α
b+α
b−α
b+α
λ
== −
)(5.0
)(5.0
)(5.0
)(5.0
0000
0000
0000
0000
0000
2221
22211
1211
1211
1
22
22
11
11
1
1
jbb
jbb
jbb
jbb
b
PBB
J
J
J
J
PAPA
[ ]222222211211121111 jCCjCCjCCjCCCCPC −+−+== −
Using theorem 6.8 the state equation is controllable if and only if
;0)(5.0;0)(5.0;0 222112111 ≠±≠±≠ jbbjbbb equivalently if and only if
2,1;0;0;0 211 =≠≠≠ iforborbb ii
The state equation is observable if and only if
;0;0;0 2221121111 ≠±≠±≠ jccjccc equivalently , if and only if
2,100;0 211 =≠≠≠ iforcorcc ii
6.17 find two- and three-dimensional state equations to describe the network shown in fig 6.12 .
discuss their controllability and observability
找到图 6.12 所 网络 两维和三维 态 述。讨论它们 可控和可观性，
solution: the network can be described by the following equation :
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32133212
1 ;;3;2
2
xyxxxxxxx
xu
=−−==−=
+
they can be arranged as
213
13
12
11
22
1
22
1
22
3
22
3
11
2
11
2
xxxy
uxx
uxx
uxx
−−==
+=
+=
−−=
so we can find a two-dimensional state equation to describe the network as
following : 213
2
1
2
1
22
3
11
2
0
22
3
0
11
2
xxxyu
x
x
x
x
−−==
−
+
−
=
and it is not controllable because 21)
22
3
11
2
22
3
11
2
11
2
()(
2
<=
−−
−
r=r c
however it is observable because 2)
0
22
1
11
()( =
−−
r=r o
we can also find a three-dimensional state equation to describe
it : [ ]xyu
x
x
x
x
x
x
100
22
1
22
3
11
2
00
22
1
00
22
3
00
11
2
3
2
1
3
2
1
=
−
+
−
=
and it is neither controllable nor
observable because [ ] 32)(31)(
2
2 <=
r<=r
CA
CA
C
BAABB
6．18 check controllability and observability of the state equation obtained in problem 2.19 . can uou
give a physical interpretation directly from the network
验 2.19 中所得 态 可控性和可观性。能否 接从该网络给出一个 解释？
Solution : the state equation obtained in problem 2.19 is
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[ ]xyu
x
x
x
x
x
x
010
0
1
1
110
100
001
3
2
1
3
2
1
=
+
−=
It is controllable but not observable because
[ ] 2
110
100
010
)(3
110
101
111
)(
2
2 ≤
−−
−r=
r=
−r=r
CA
CA
C
BAABB
A physical interpretation I can give is that from the network , we can that if 0)0(0 1 ≠== axu
and 0)0(2 =x then 0≡y that is any [ ]Tax *0)0( = and 0)( ≡tu yield the same output
0)( ≡tY thus there is no way to determine the initial state uniquely and the state equation describing
the network is not observable and the input is has effect on all of the three variables . so the state
equation describing the network is controllable
6.19 continuous-time state equation in problem 4.2 and its discretized equations in problem 4.3 with
sampling period T=1 and π . Discuss controllability and observability of the discretized equations .
考虑 4.2 中 连续 态以及 4.3 中它 化 态 （T=1 and π）. 讨论 态
可控性和可观性
solution : the continuous-time state equation in problem 4.2 is [ ]xyuxx 32
1
1
22
10
=
+
−−
=
the discretized
equation :
[ ] [ ] [ ]ku
econTT
eTconT
kx
eTTTe
TeeTT
kx
T
T
TT
TT
++
+−+
−−
+
=+
−
−
−−
−−
)sin2(1
)sin
2
1
2
3
(
2
3
)sin(cossin2
sin)sin(cos
1
[ ] [ ] [ ]kxky 32=
T=1
−
=
−−
=
1821.0
0471.1
1108.06191.0
3096.05083.0
dd BA
T= π :
−
=
=
0432.1
5648.1
0432.00
00432.0
dd BA
The xontinuous-time state equation is controllable and observable , and the sustem it describes is
single-input so the drscretized equation is controllable and observable if and only if
[ ] 2,12/2|| =π≠λ−λ mformjI im whenever [ ] 0=λ−λ jR ie the eigenvalues of a are
–1+j and –1-j . so [ ] 2|| =λ−λ jI im for T=1 . 2,122 =≠π manyform .so the
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discretized equation is neither controllable nor observable . for T= π 2m=1 .so the descretized
equation is controllable and observable .
6.20 check controllability and observabrlity of [ ]xyux
t
x 10
1
0
0
10
=
+
= 验可控性和
观性。
Solution :
−
−=
t
MtAM
1
)( 01 the determinant of the matrix [ ]
−
−
=
t
MM
1
10
10 is 1. which
is nonzero for all t . thus the state equation is controllable at every t。
Its state transition matrix is
τ−
=Φ
−
τ∫
)(5.0
5.05.0
0
2
0
2
0
22
0
1
),(
tt
t
t
t
e
dee
tt and
[ ] [ ])202
)2
0
2
0
22
(5.0
(5.0
5.05.0
0 0
0
1
10),()(
t
t
t
t e
e
dee
ttC
−τ
−τ
ττ
=
ϑ−
=Φτ ∫
[ ]
τ
=
τ⋅
=
∫
∫
−τ
−τ
−τ
d
e
de
e
ttW
t
t t
t
t
t t
1
0
2
0
2
2
0
21
0
2
0
2
0
00
0
0
),(
)(5.0
)(5.0100
We see . ),( 100 ttW is singular for all 0t and t , thus the state equation is not observable at any0t .
6.21 check controllability and observability of [ ]xeyu
e
xx t
t
−
− =
+
−
= 0
1
10
00
solution :
textxxtx −== )0()()0()( 2211 its state transition matrix is
=
==Φ +−−
−
00 0
01
0
01
0
01
)(),( 0
1
0 tttt eee
tXtXtt ）（ and
[ ] [ ]0
0
2
0 0
0
01
0),()(
11
0
01
)(),(
t
t
t
tt
e
e
etC
eee
Bt
+τ−
+τ−
−
−τ−τ+−
=
=τΦτ
=
=ττΦ
WE compute [ ]
−−
−−
=τ
=τ
=
−−
−
−−
−
−
− ∫∫
)()(
)(1
1
1
),(
0
2
0
00
20
00 ttette
ttett
d
ee
e
de
e
ttW
tt
t
t
t tt
t
t
t
t t
c
Its determinant is identically zero for all to and 0t , so the state equation is not controllable at any 0t
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[ ] τ
=τ
= ∫∫ +τ−+τ−+τ− dedeettW
t
t t
t
t
t t
c
0
0
0
0
0 24
2
20
0
00
1
1
),( its determinant is identically zero for all
0t and 1t , so the state equation is not observable at any 0t .
6.22 show shat ( ))(),(( tBtA is controllable at 0t if and only if ))(),(( tBtA ′′− is observable at 0t
证 ))(),(( tBtA 在 0t 可控当且仅当 ))(),(( tBtA ′′− 在 0t 可观。
Proof: Itttt =ΦΦ ),(),( 00 where ),( 0ttΦ is the transition matrix of ))()()( txtAtx =
0),(),()(),(),(),(),()(
),(
),(),(
),(
),(),(
000000
0
00
0
00
=Φ
∂
∂
Φ+=Φ
∂
∂
Φ+ΦΦ=
∂
Φ∂
Φ+Φ
∂
Φ∂
=ΦΦ
∂
∂
∴
tt
t
tttAtt
t
tttttttA
t
tt
tttt
t
tt
tttt
t
There we have )(),(
),(
0
0 tAtt
t
tt
+Φ−=
∂
Φ∂
and. ),()(
),(
0
0 tttA
t
tt
Φ′+′−=
∂
Φ∂
So ),( 0 ttΦ′ is the transition matrix of ))()()( txtAtx ′= ))(),(( tBtA is controllable at 0t if and only
if there exists a finite 01 tt > such that ττΦ′τ′ττΦ= ∫ dtBBtttW
t
t
c ),()()(),(),( 110
1
0
is nonsingular
))(),(( tBtA ′′− is observable at 0t if and only if there exists a finite 01 tt > such that
ττΦ′τ′ττΦ= ∫ dtBBtttW
t
t
c ),()()(),(),( 000
1
0
is nonsingular.
For time-invariant systems , show that （A,B）is controllable if and only if (-A,B) is controllable , Is true
foe time-varying systems ?
证 对 不变 统，（A,B）可控当且仅当（-A,B）可控。 对 变 统 否 同结论？
Proof: for time-invariant system (A,B) is controllable if and only if
[ ] nBAABBC n =r=r −11)( (assuming a is nn×
(-A,B) is controllable if and only if [ ] nBAABBC n =r=r −12 )(
we know that any column of a matrix multiplied by nonzero constant does not change the rank of the
matrix . so =r )( 1C )( 2Cr the conditions are identically the same , thus we conclude that (A,B)is
controllable if and only if (-A,B)is controllable .for time-systems, this is not true .
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7.1 Given
）（ 2)(1
1
)(
2 +−
−
=
ss
s
sg . Find a three-dimensional controllable realization check its
observalility
找
）（ 2)(1
1
)(
2 +−
−
=
ss
s
sg 一三维可控性 ，判 其可观性。
Solution :
）（）（ 22
1
2)(1
1
)(
222 −−+
−
+−
−
=
sss
s
ss
s
sg using (7.9) we can find a
three-dimensional controllable realization as following
[ ]xyuxx 110
0
0
1
010
001
212
−=
+
−
= this realization is not observable because
(s-1)and ）（ 2)(12 +− ss are not coprime
7.2 find a three-dimensional observable realization for the transfer function in problem 7.1
check its controllability
找 7.1 中
）（ 2)(1
1
)(
2 +−
−
=
ss
s
sg 一三维可控性 ，判 其可观性。
Solution : using (7.14) ，we can find a 3-dimensional observable realization for the transfer
function : [ ]xyuxx 001
1
1
0
002
101
012
=
−
+
−
= and this tealization is not
controllable because (s-1) and ）（ 2)(12 +− ss are not coprime
7.3 Find an uncontrollable and unobservable realization for the transfer function in
problem 7.1 find also a minimal realization
7.1 中传函 一个不可控且不可观 以及一个 小 。
Solution :
23
1
)2)(1(
1
2)(1
1
)(
22 ++
=
++
=
+−
−
=
ssssss
s
sg
）（
a minimal realization is ,
[ ]xyuxx 10
0
1
01
23
=
+
−−
= an uncontrollable and unobservable realization
is [ ]xyuxx 010
0
0
1
100
001
023
=
+
−−
=
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7．4 use the Sylvester resultant to find the degree of the transfer function in problem 7.1
Sylvester resultant 7.1 中传函
solution :
−
−−−α
−−−
−−
=
010000
020100
110201
12110
001211
000012
s
rank(s)=5. because all three D-columns of s are linearly independent . we conclude that s has only
two linearly independent N-columns . thus deg 2)(ˆ =sy
7.5 use the Sylvester resultant to reduce
14
12
2 −
−
s
s
to a coprime fraction Sylvester resultant
将
14
12
2 −
−
s
s
化 为 分式。
Solution : ssusranks ===
−−
−−
= 11,3)(
0400
2004
1120
0011
null
T
s
−= 10
2
1
2
1
)( 1 thus we have ssDssN +=⋅+=
2
1
)(0
2
1
)( and
12
1
2
1
2
1
14
12
2 +
=
+
=
−
−
s
s
s
s
7.6 form the Sylvester resultant of
）（ ss
s
sg
2
2
)(ˆ
2 +
+
= by arranging the coefficients of
)(sN and )(sD in descending powers of s and then search linearly independent columns in
order from left to right . is it true that all D-columns are linearly independent of their LHS
columns ?is it true that the degree of )(ˆ sg equals the number of linearly independent N-columns?
）（ ss
s
sg
2
2
)(
2 +
+
= 将 )(sN 和 )(sD 按 s 幂排列形成 Sylvester resultant ， 后从左
至右找出 性 关列。所 D-列 否都与其 LHS 性 关？ )(ˆ sg 否 于 性
关 N-列 回？
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Solution ；
)(
)(
:
2
2
)(ˆ
2 sN
sD
ss
s
sg =
+
+
=
）（
1)(ˆdeg =sg
01
2
2
01
2
2
)(
)(
NSNSNsN
DSDSDsD
++=
++=
from the Sylvester resultant :
=
2000
1220
0112
0001
s 23)( == usrank (the number of linearly independent
N-columns
7.7 consider
)(
)(
)(ˆ
21
2
21
sN
sD
ss
s
sg =
∂+α+
b+b
=
）（
and its realization
[ ]xyuxx 2121
0
1
01
bb=
+
α−α−
=
show that the state equation is observable if and only if the Sylvester resultant of
)()( sNandsD is nonsingular . 考 虑
)(
)(
)(ˆ
21
2
21
sN
sD
ss
s
sg =
∂+α+
b+b
=
）（
及 其 实
[ ]xyuxx 2121
0
1
01
bb=
+
α−α−
= 证 该 态 可 观 及 其 当 且 仅 当
)()( sNandsD Sylvester resultant 奇异
proof: [ ]xyuxx 2121
0
1
01
bb=
+
α−α−
= is a controllable canonical form realization of
)(ˆ sg from theorem 7.1 , we know the state equation is observable if and only id )()( sNandsD
are coprime
from the formation of Sylvester resultant we can conclude that )()( sNandsD are coprime if
and only if the Sylvester resultant is nonsingular thus the state equation is observable if and only if the
Sylvester resultant of )()( sNandsD is nonsingular
7.8 repeat problem 7.7 for a transfer function of degree 3 and its controllable-form realization , 对一
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个 3 传函及其可控型实 重复 7.7
solution : a transfer function of degree 3 can be expressed as
)(
)(
)(ˆ
32
2
1
3
32
2
1
3
1
sD
sN
sss
ssssg =
α+∂+α+
b+b+b+b
=
）（
where )()( sNandsD are coprime
writing )(ˆ sg as
）
（
32
2
1
3
032022
2
010 )())(ˆ
α+α+α+
bα+b+bα−b+bα−b
=
sss
ss
sg , we can obtain a controllable
canonical form realization of )(ˆ sg
[ ] uxy
uxx
0033022011
321
0
0
1
010
001
b+bα−bbα−bbα−b=
+
α−α−α−
=
the Sylvester resultant of )()( sNandsD is nonsingular because )()( sNandsD are
coprime , thus the state equation is observable .
7.9 verify theorem 7.7 fir
2)1(
1
)(ˆ
+
=
s
sg .对
2)1(
1
)(ˆ
+
=
s
sg 验定定 7.7
verification :
2)1(
1
)(ˆ
+
=
s
sg is a strictly proper rational function with degree 2, and it can be
expanded into an infinite power series as
++−+−+⋅= −−−−−− 654321 54320)(ˆ sssssssg 1−s
[ ]( )
( ) 2,1,.22,2)2,2(
1)1,2()2,1(00)1,1(
==++=
====
lkveyerforLKTT
TTT
rr
rrrr
7.10 use the Markov parameters of
2)1(
1
)(ˆ
+
=
s
sg to find on irreducible companion-form realization
马尔 夫参
2)1(
1
)(ˆ
+
=
s
sg 一个不可 型实
solution
2
21
10
)2,2(
5)4(3)2(0)(ˆ 654321
=
−
=
++−++−++⋅= −−−−−−
rrT
sssssssg
deg )(ˆ sg =2
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[ ]01
1
0
21
10
01
12
32
21
21
10
32
21
)2,2()2,2(
~
1
1
=
=
−−
=
−
−
=
−
−
−
==∴
−
−
cb
TTA
the triplet ),,( cbA is an irreducible com-panion-form realization
7.11 use the Markov parameters of
2)1(
1
)(ˆ
+
=
s
sg to find an irreducible balanced-form realization
)(ˆ sg 马尔 夫参 它 一个不可 均衡型实 .
Solution :
−
=
21
10
)2,2(T
Using Matlab I type
[ ] [ ]
[ ]
)1,1(),1,(;
)(**)(
;*;*
);(),(,,
32;2121;10
Occb
cinvoinvA
lslcslko
ssqrtsltsvdlsk
ttt
==
=
′==
==
−−=−=
ϑ
This yields the following balanced realization
[ ]xy
xx
5946.05946.0
5946.0
5946.0
2929.07071.0
7071.07071.1
−−=
−
+
−−
−
=
7.12
Show that the two state equation [ ]xyuxx 22
0
1
10
12
=
+
= and
[ ]xyuxx 02
2
1
11
02
=
+
−−
= are realizations of )
2(
)22(
2 −−
+
ss
s
, are they
minimal realizations ? are they algebraically equivalent?
证 以上两 态 都 )
2(
)22(
2 −−
+
ss
s
,他们 否 小 ? 否代 价?
Proof :
2
2
)2)(1(
)1(2
)2(
)22(
2 −
=
−+
+
=
−−
+
sss
s
ss
s
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[ ] [ ]
[ ] [ ]
2
2
2
1
1
1
)2)(1(
1
0
2
1
02
2
1
11
02
02
2
2
0
1
1
1
0
)2)(1(
1
2
1
22
0
1
10
12
22
1
1
−
=
+−+
−
−=
+
−
−
=
−
−−−=
−
−−
−
−
s
sss
s
s
s
s
s
sss
s
s
thus the two state equations are realizations of )
2(
)22(
2 −−
+
ss
s
the degree of the transfer function
is 1 , and the two state equations are both two-dimensional .so they are nor minimal realizations .
they are not algebraically equivalent because there does not exist a nonsingular matrix p such that
[ ] [ ]0222
0
1
0
1
11
02
10
12
1
1
=
=
−−
=
−
−
P
P
PP
7.13 find the characteristic polynomials and degrees of the following proper rational matrices
++
+
+
=
13
1
1
31
ˆ
1
s
s
s
s
s
sG
+++
+++=
)2)(1(
1
2
1
)2)(1(
1
)1(
1
ˆ
2
1
sss
sss
G
+
+
+
+
+
+
+=
s
s
s
s
s
s
s
sG
1
5
1
4
1
2
1
2
3
)1(
1
ˆ
2
3
note that each entry of )(ˆ sGs has different poles from other entries
则 )(ˆ1 sG , )(
ˆ
2 sG 和 )(
ˆ
3 sG 征多 式和 .
Solution : the matrix )(ˆ1 sG has
1
,
3
1
,
1
3
,
1
+++
+
s
s
ss
s
s
and det )(ˆ1 sG =0 as its minors
thus the characteristic polynomial of )(ˆ1 sG is s(s+1)(s+3) and )(
ˆ
1 sG =3 the matrix )(
ˆ
2 sG has
2)1(
1
+s
,
)2)(1(
1
++ ss
,
2
1
+s
,
)2)(1(
1
++ ss
. det
23
2
2
)2()1(
1
)(ˆ
++
+−−
=
ss
ss
sG as
its minors , thus the characteristic polynomial of 5)(ˆ)2()1()(ˆ 2
23
2 =++= sGandsssG d
Every entry of )(ˆ 3 sG has poles that differ from those of all other entries , so the characteristic
polynomial of )(ˆ 3 sG is )5)(4()3)(2()1(
22 +++++ ssssss and 8)(ˆ 3 =sGd
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7.14 use the left fraction
−
−
=
−
1
11
)(ˆ
1
ss
s
sG to form a generalized resultant as in (7.83), and
then search its linearly independent columns in order from left to right ,what is the number of linearly
independent N-columns ?what is the degree of )(ˆ sG ? Find a right coprime fraction lf )(ˆ sG ,is the
given left fraction coprime? )(ˆ sG 左分式 成(7.83)中所 广义终结 , 后从左到右找出其
性 关列. 性 关 N-列 多少> )(ˆ sG 多少? 出 )(ˆ sG 一个 右分式.
给 左分式 否 ?
Solution : )()(:
1
11
)(ˆ 1
1
sNsD
ss
s
sG −
−
=
−
−
=
Thus we have
−
=
−
+
=
−
=
1
1
)(
11
01
00
101
)(
sN
s
ss
s
sD
And the generalized resultant
0 1 1 0 0 0
0 0 -1 0 0 0
1 0 0 0 1 1
-1 1 0 0 0 -1
0 0 0 1 0 0
0 0 0 -1 1 0
S
=
rank s=5 ,
the number of linearly independent N-colums is l. that is u=1,
[ ]
1100
100001)(
DNDN
snull
−−
−=
So we have
1)(ˆdeg
0
1
)(ˆ
0
1
0
0
0
1
)(
10)(
1
===
=
+
=
+
+
=
=⋅+=
−
usG
ssG
sSN
SSSD
the given left fraction is not coprime .
7.15 are all D-columns in the generalized resultant in problem 7.14 linearly independent .pf their LHS
columns ?Now in forming the generalized resultant ,the coefficient matrices of )(SD and )(SN are
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arranged in descending powers of s , instead of ascending powers of s as in problem 7.14 .is it true that
all D-columns are linearly independent of their LHS columns? Doe’s he degree of )(ˆ sG equal the
number of linearly independent N-columns ?does theorem 7.M4hold ? 7.14 中 广义终结 中 否
所 D-列都与其 LHS 列 性 关? 将 )(SD 和 )(SN 按 S得 幂.排列 成另一
形式 广义终结 .在这 终结 中 否所 D-列也与其 LHS 列 性 关? )(ˆ sG
否 于 性 关 N-列 ?定 7.M4 否成 ?
Solution : because ,01 ≠D ALL THE D-columns in the generalized resultant in problem7.14 are
linearly independent of their LHS columns
Now forming the generalized resultant by arranging the coefficient matrices of )(SD and )(SN in
descending powers of s :
5
100000
110000
011100
001110
000011
000001
00
00
00
1100
11
=
−
−−
−
=
= Srank
ND
NDND
ND
S
we see the 0D -column in the second colums block is linearly dependent of its LHS columns .so it is
not true that all D-columns are linearly independent of their LHS columns .
the number of linearly independent N -columns is 2 and the degree of )(ˆ sG is I as known in problem
7.14 , so the degree of )(ˆ sG does not equal the number of linearly independent N -columns , and the
theorem 7.M4 does not hold .
7.16, use the right coprime fraction of )(ˆ sG obtained in problem 7.14 to form a generalized tesultant as
in (7.89). search its linearly independent rows in order from top to bottom , and then find a left coprime
fraction of )(ˆ sG 7.14 中得到 )(ˆ sG 右分式,如式(7.89) 造一个广义终结 , 从上至
下找出其性 关行, 出 )(ˆ sG 一个 左分式.
Solution :
==
= −
0
1
)()(
0
1
)(ˆ 1 sNssDssG
The generalized resultant
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0,13
000
010
100
000
001
010
21 ===
= γγTrankT
[ ]
[ ]′−=
′=
1001)
010
100
001
010
(
100)
000
001
010
(
mull
mull
[ ]
+
=
=
+
=
−
=−−∴
0
1
)(
10
0
00
01
10
00
)(
000100
010001
0000
sN
s
ssD
DNDN
thus a left coprime fraction of
−−
+
5.05.0
5.05.02
s
sss
is
=
−
0
1
10
0
)(ˆ
1
s
sG
7.17 find a right coprime fraction of
+
++
=
ss
s
s
s
s
s
sG
22
121
)(ˆ
2
23
2
snd then a minimal realization
)(ˆ sG 一个 右分式及一个 小 ]
solution : )()(:
22
121
)(ˆ 1
2
23
2
SNSD
ss
s
s
s
s
s
sG −=
+
++
=
where
2
2
32
2
3
00
21
21
10
02
01
22
)12(1
)(
00
01
10
00
00
00
00
00
0
0
)(
ss
ss
sss
sN
sss
s
s
sD
+
+
=
+
++
=
+
+
+
=
=
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the generalized resultant is
=
000000000000
000100000000
001000000000
210000010000
210000100000
100021000001
020021000010
010010002100
000002002100
00001001000
000000000200
000000000100
S
rank 1,2,9 21 === µµs
the monic null vectors of the submatrices that consist of the primary dependent 2N -columns and
1N -columns are , respectively
[ ]
[ ]
′
−−
−−−
−−−
=
−
−
∴
−−−−−=
′−−−=
000015.000
010005.011
5.005.25.2
5.005.25.0
1005.0115.005.25.01
15.0005.005.25.22
2
2
1
1
0
0
D
N
D
N
D
N
Z
Z
2
2
0 0 0.5 0.5 1 0 0.5 0.5
( )
0.5 0.5 0 1 0 0 0.5 0.5
0.5 2.5 1 0 0.5 2.5
( )
2.5 2.5 1 0 2.5 2.5
s s s
D s s s
s
s
N s s
s
+
= + + = − − − −
+
= + = +
thus a right coprime fraction of )(ˆ sG is
−−
+
+
+
=
5.05.0
5.05.0
5.25.2
5.25.0
)(ˆ
2
s
sss
s
s
sG
we define
=
=
10
01
0
)(
0
0
)(
2
s
sL
s
s
sH
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then we have
1
1
1
1 0.5 0.5 0 0
( ) ( ) ( )
0 1 0 0.5 0.5
1 0.5 2.5
( ) ( )
1 2.5 2.5
1 0.5 1 0.5
0 1 0 1
1 0.5 0.5 0 0 0.5 0.25 0.25
0 1 0 0.5 0.5 0 0.5 0.5
hc
hc ic
D s H s L s
N s L s
D
D D
−
−
−
= + − −
=
−
= =
−
= = − − − −
thus a minimal realization of )(ˆ sG is
0.5 0.25 0.25 1 0.5
1 0 0 0 0
0 0.5 0.5 0 1
1 0.5 2.5
1 2.5 2.5
x x u
y x
− − − −
= +
=
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8.1 given [ ]xyuxx 11
2
1
11
12
=
+
−
=
find the state feedback gain k
so that the feedback system has –1 and –2 as its eigenvalues .
compute
k
directly with using any equivalence transformation 对 给 态 , 不 价交换,
接 态反馈增 k 使 态反馈 统 征值为-1,-2.
Solution : introducing state feedback
[ ]xkkru 21−= , we can obtain
[ ] rxkkxx
+
−
−
=
2
1
2
1
11
12
21
the new A-matrix has characteristic polynomial
)35()32(
)1)(21()21)(2()(
2121
2
2121
+−+−++=
−−−−+−+−=∆
kkskks
kkkskssf
the desired characteristic polynomial is 23)2)(1( 2 ++=++ ssss ,equating
332 21 =−+ kk and 235 21 =+− kk , yields
14 21 == kandk so the desired state feedback gain k is [ ]14=k
8.2 repeat problem 8.1 by using (8.13), 利 式(8.13)重解 8.1.
solution :
[ ] [ ]
[ ] trollablecanisbAsogularnonisCCbAbD
CC
k
sssss
sssss
f
),(,sin
7
1
7
2
7
4
7
1
12
41
10
31
10
31
1632)3(3
23)2)(1()(
131)1)(2()(
1
1
2
2
−
−
=
==
=
−
=
−=−−−=
++=++=∆
+−=+−−=∆
−
−
using (8.13)
[ ] [ ]14
7
1
7
2
7
4
7
1
10
31
161 =
−
−
−== −CCkk
8.3 Repeat problem 8.1 by solving a lyapunov equation
解 lyapunov 重解 8.1.
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solution : ),( bA is controllable
selecting [ ]11
20
01
=
−
−
= kandF then
[ ] [ ]14
013
19
11
013
19
13
9
1
13
1
0
1
1
=
−
==
−
=
=⇒=−
−
−
Tkk
TTkbTFAT
8.4 find the state foodback gain for the state equation uxx
+
−
=
1
0
1
100
110
211
so that the resulting system has eigenvalues –2 and –1+j1 . use the method you think is the
simplest by hand to carry out the design 态反馈增 . 使 给 态 态反馈 统具
-2 和-1+j1 征值/
solution : ),( bA is controllable
[ ] [ ]
−
−−=
−−
=
=
−
−
=
=−−−−−=
+++=−++++=∆
−+−=−=∆
−
−
121
232
011
111
210
211
100
310
631
100
310
331
537)1(436)3(4
464)11)(11)(2()(
133)1()(
1
1
23
233
CC
CC
k
sssjsjsss
sssss
f
Using(8.13)
[ ] [ ]84715
121
232
011
100
310
631
5371 −=
−
−−
= −CCkk
8.4 Consider a system with transfer function
)3)(2)(1(
)2)(1(
)(ˆ
+−+
+−
=
sss
ss
sg is it possible to
change the transfer function to
)3)(2(
)1(
)(ˆ
++
−
=
ss
s
sg f by state feedback? Is the resulting
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system BIBO stable ?asymptotically stable?
否能够通过 态反馈将传函 )(ˆ sg 变为 )(ˆ sg f ?所得 统 否 BIBO 定? 否逐 定?
Solution :
)3()2
)2)(1(
)3)(2(
)1(
)(ˆ
2 ++
+−
=
++
−
=
ss
ss
ss
s
sg f
We can easily see that it is possible to change )(ˆ sg to )(ˆ sg f by state feedback and the resulting
system is asymptotically stable and BIBO stable .
8.6 consider a system with transfer function
)3)(2)(1(
)2)(1(
)(ˆ
+−+
+−
=
sss
ss
sg is it possible to
change the transfer function to
3
1
)(ˆ
+
=
s
sg f by state feedback ? is the resulting system BIBO
stable? Asymptotically stable ?
能否通过 态反馈将传函 )(ˆ sg 变为 )(ˆ sg f ?所得 统 否 BIBO 定? 否逐 定?
Solution ;
)3)(2)(1(
)2)(1(
3
1
)(ˆ
++−
+−
=
+
=
sss
ss
s
sg f
It is possible to change )(ˆ sg to )(ˆ sg f by state feedback and the resulting system is
asymptotically stable and BIBO stable .. however it is not asymptotically stable .
8.7 consider the continuous-time state equation
[ ]xyuxx 002
1
0
1
100
110
211
=
+
−
=
let xkpru −= , find the feedforward gain p and state feedback gain k so that the resulting
system has eigenvalues –2 and 11 j±− and will track asymptotically any step reference input
对 给连续 态 , 令 xkpru −= 前馈增 p 和 态反馈增 k 使所得 统
征值为-2 和 11 j±− , 并且能够 进跟踪任意 跃参考输入.
Solution:
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[ ]
[ ]84715
5.0
8
4
464)(
133
882
1
0
1
1
1
00
)1(
1
1
1
0
)1(
1
)1(
1
)1(
1
1
1
002)()(ˆ
3
3
23
23
2
2
232
−=
===∴
+−++=∆
−+−
+−
=
−
−−
−
−
−−−
=−=
k
P
ssss
sss
ss
s
ss
ssss
bAsIcsg
f
b
α
(obtained at problem 8.4)
8.8 Consider the discrete-time state equation
[ ] ][002][
][
1
0
1
100
110
211
]1[
kxky
kuxkx
=
+
−
=+
find the state feedback gain so that the resulting system has all eigenvalues at x=0 show that for
any initial state the zero-input response of the feedback system becomes identically zero for
3≥k
对 给 态 , 态反馈增 使所得 统所 征值都在 Z=0, 证 对任意
初始 态,反馈 统 输入响应均为 ( 3≥k )
solution ; ),( bA is controllable
[ ]
[ ] [ ]251
121
232
011
100
310
631
133
133
)(
133)(
1
3
3
=
−
−−
−==
−=
=∆
−+−=∆
− kCCk
k
zz
zszz
f
The state feedback equation becomes
[ ]
[ ] ][002][
][
1
0
1
][
151
110
440
][
1
0
1
][251
1
0
1
100
110
211
]1[
kxky
krkxkrkxkx
=
+
−−−
−−
=
+
−
−
=+
denoting the mew-A matrix as A , we can present the zero-mput response of the feedback system
as following ]0[][ xACky
k
zi =
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compute
k
A
1
1
000
100
010
000
100
010
−
−
=
=
QQA
QQA
K
k
using the nilpotent property
30
000
100
010
≥=
kfor
K
so we can readily obtain
0]0[0][ == xCky zi for 3≥k
consider the discret-time state equation in problem 8.8 let for ][][][ kxkkprku −= where p
is a feedforward gain for the k in problem 8.8 find a gain p so that the output will track any step
reference input , show also that y[k]=r[k] for 3≥k ,thus exact tracking is achieved in a finite
number of sampling periods instead of asymptotically . this is possible if all poles of the resulting
system are placed at z=0 ,this is called the dead-beat design
对 8.8中 态 ,令 ][][][ kxkkprku −= ,其中 p 前馈增 ,对 8.8中所得
k , 一增 p 使输出能够跟踪任意 跃参考输入,并证 3≥k y[k]=r[k].实 跟踪 在
多采 周 中得到 ,而不 进 . 当反馈 统 所 都被配置在 Z=0 ,这
跟踪 可以做到 ,这就 所谓 小[拍设计.
Solution ;
3
2
3
23
2
88
)(ˆ
][][][)(
133
882
)(ˆ
z
zz
pzg
kxkkprkuzz
zsz
sz
zg
f
+−
=
−==∆
−+−
+−
=
),( bA is controllable all poles of )(ˆ zg f can be assigned to lie inside the section in fig 8.3(b)
under this condition , if the reference input is a step function with magnitude a , then the output
y[k] will approach the constant +∞→⋅ kasag f )1(ˆ
thus in order for y[k] to track any step reference input we need
1)1(ˆ =fg
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5.012)1(ˆ =⇒== ppg f
the resulting system can be described as
[ ] ][002][
][][
][
5.0
0
5.0
][
151
110
440
][
kxky
krbkxA
krkxkx
=
+=
+
−−−
−−
=
the response excited by r[k] is
][]0[][
1
0
1
mrbACxACky
K
M
mKK
∑
−
=
−−
+=
sokforAknowweas
K
,30, ≥=
3]3[4]2[4]1[
]3[]2[]1[][
2
≥−+−−−=
−+−+−=
kforkrkrkr
krbAckrbAckrbcky
for any step reference input r[k]=a the response is
y[k]=(1-4+4) a =a =r[k] for 3≥k
8.10 consider the uncontrollable state equation uxx
+
−
−
=
1
1
1
0
1000
0100
0020
0012
is it possible
to find a gain k so that the equation with state feedback xkru −= has eigenvalues –2, -2 ,
-1 ,-1 ? is it possible to have eigenvalues –2, -2 ,–2 ,-1 ? how about –2 ,-2, –2 ,-2 ? is the equation
stabilizable?对 给不可控 态 , 否可以 出一增 k 使具 反馈 xkru −= 具
征值-2. –2, -1, -1, ? 征值-2, -2, -2,-1? 征值-2,-2,-2,-2?该 否 定?
Solution: the uncontrollable state equation can be transformed into
u
b
x
x
A
A
u
x
x
x
x
c
c
c
c
c
c
c
c
c
+
=
+
−
−
=
00
0
0
0
0
1
1000
0310
0001
0400
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Where the transformation matrix is
−
−
=−
1111
0111
0421
0410
1P
),( cC bA is controllable and )( CA is stable , so the equation is stabilizable . the eigenvalue of
)(
C
A ,say –1 , is not affected by the state feedback ,while the eigenvalues of the controllable
sub-state equation can be arbitrarily assigned in pairs .
so by using state feedback ,it is possible for the resulting system to have eigenvalues –2 ,-2 , -1, -1,
also eigenvalues –2, -2, -2, -1.
It is impossible to have –2, -2, -2, -2, as it eigenvalues because state feedback can not affect the
eigenvalue –1.
8.11 design a full-dimensional and a reduced-dimensional state estimator for the state equation in
problem 8.1 select the eigenbalues of the estimators from { }223 j±−−
对 8.1 中 态 , 设计一全维 态估计器和一将维估计器.估计器 征值从
{ }223 j±−− 中选取.
Solution : the eigenvalues of the full-dimensional state estimator must be selected as 22 j±− ,
the A-, −b and −c matxices in problem 8.1 are
[ ]11
2
1
11
12
=
=
−
= cbA
the full-dimensional state estimator can be written as ylubxclAx ++=−= ˆ)(̂
let
=
2
1
l
l
l then the characteristic polynomial of )(ˆ clAx −= is
1912
844)2(
23)3(
)1)(1()1)(2()(
21
22
2121
2
1221
=−=⇒
++=++=
−−+−++=
−+++−+−=∆′
ll
sss
llslls
lllslss
thus a full-dimensional state estimor with eigenvalues 22 j±− has been designed as following
yuxx
−
+
+
−−
=
19
12
2
1
ˆ
1820
1314̂
designing a reduced-dimensional state estimator with eigenvalue –3 :
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(1) select |x| stable matrix F=-3
(2) select |x| vector L=1, (F,L) IS controllable
(3) solve T=: TA-FT= cl
let T :
[ ]
[ ] [ ] [ ]
=⇒
=+
−
=
21
4
21
3
113
11
12
2121
21
T
tttt
ttT
(4) THE l-dimensional state estimutor with eigenvalue –3 is
−
−
=
=
++−=
−
z
y
z
y
x
yuxz
215
214
21
4
21
5
11
ˆ
21
13
3
1
8.12 consider the state equation in problem 8.1 .compute the transfer function from r to y of the
state feedback system compute the transfer function from r to y if the feedback gain is applied to
the estimated state of the full-dimensional estimator designed in problem 8.11 compute the
transfer function from r to y if the feedback gain is applied to the estimated state of the
reduced-dimensional state estimator also designed in problem 8.11 are the three overall transfer
functions the same ?
对 8.1 中 态 ,计 态反馈 统从 r 到 y 传送函 .若将反馈增 应 到 8.11
中所设计 全维 态估计器所估计 估计 态,计 从 r 到 y 传送函 ,若将反馈增 应
到 8.11 中所设计 维 态估计器 估计 态, 计 从 r 到 y 传送函 , 这三个传送函
否 同?
Solution
(1)
[ ]
)2)(1(
43
2
1
1
1
)2)(1(
9
0
2
1
11
)(ˆ 1
++
−
=
+++
+=
+−=∴
−==+=
−
→
ss
s
sss
s
bkbAsIcy
xkruxcyubxAx
yx
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(2)
[ ]
[ ] [ ]
[ ]xy
rxx
xxrx
yuxx
rxx
xrxuxx
21
2
1
1919
1212
ˆ
2028
1210
11
19
12
ˆ14
2
1
2
1
ˆ
1820
1314
19
12
2
1
ˆ
1820
1314
ˆ
2
1
ˆ
28
14
11
12
ˆ14
2
1
2
1
11
12
2
1
11
12
=
+
−−
+
−−
=
−
+
−
+
−−
=
−
+
+
−−
=
+
−
−
=
−
+
−
=
+
−
=
[ ]
=
+
−−
−−
−−−
−−
=
∴
x
x
y
r
x
x
x
x
ˆ
0011
2
1
2
1
ˆ
20281919
12101212
2811
1412
̂
[ ]
)2)(1(
43
)84)(2)(1(
)84)(43(
1632227
32883
2
1
2
1
20281919
12101212
2811
1412
0011ˆ
2
2
234
23
1
++
−
=
++++
++−
=
++++
−++
=
+−
−−
−−
−−−
=∴
−
→
ss
s
ssss
sss
ssss
sss
s
s
s
g yr
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(3)
[ ]
[ ] [ ]
[ ]xy
rzx
x
z
y
rzz
rzx
rzxx
r
z
y
xx
rrxxuxx
11
21
13
42
21
164
21
164
11)
215
214
14(
21
13
3
2
1
126
63
2321
1213
2
1
126
63
11
22
11
11
12
2
1
215
214
ˆ
28
14
11
12
2
1
2
1
ˆ
28
14
11
12
2
1
11
12
=
+−
=
+
−
−
−+−=
+
+
=
+
−
−
−
−
−
=
+
−
−
−
−
=
−
+
−
−
=
+
−
=
[ ]
[ ]
)2)(1(
43
)3)(2)(1(
)3)(43(
6116
1253
21
13
2
1
42
21
164
21
164
1262321
632113
011)(ˆ
011
21
13
2
1
42
21
164
21
164
1262321
631213
23
2
1
++
−
=
+++
+−
=
+++
−+
=
+
+−−
−
−−
=∴
=
+
−
−
−
=
∴
−
→
ss
s
sss
ss
sss
ss
s
s
s
sg
z
x
y
r
z
x
z
x
yr
these three overall transfer functions are the same , which verifies the result discussed in section
8.5.
8.13 let
=
−
=
20
21
00
00
0012
3213
0100
0010
BA find two different constant matrices k such that
(A-BK) has eigenvalues j34 ±− and j45 ±− 两个不同 常 使(A-BK)具 征值
j34 ±− 和 j45 ±−
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solution : (1) select a 44× matrix
−
−−
−
−−
=
5400
4500
0043
0034
F
the eigenvalues of F are j34 ±− and j45 ±−
(2)if ),(,
0100
0001
11 kFk
= is observable
−−−−
==
−−
−−
−−
−−−
=⇒=−
−
2253.28747.141758.1190670.371
0000.22000.140000.1682000.606
1982.02451.00043.00006.0
0185.01731.00714.01322.0
0191.00193.00273.00126.0
0042.00005.00059.00013.0
1
111
1111
TKK
TKBFTAT
(3)IF
=
0100
1111
2K ),( 2kF is observable
−−−−
==
−
−−
−−−
=⇒=−
−
5853.24593.78815.595527.185
2509.30893.02145.559824.252
2007.02473.00037.00048.0
0245.03440.00607.02036.0
0306.00443.00147.00399.0
0081.00024.00071.00046.0
1
222
2222
TKK
TKBFTAT
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线性系统课后答案第2章
线性系统课后答案第3章
线性系统课后答案第4章
线性系统课后答案第5章
线性系统课后答案第6章
线性系统课后答案第7章
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